Answer:
Concentration of MgSO4 = 0.0521 × 10⁶ ppmConcentration of MgSO4 = 6% w/vConcentration of MgSO4 = 5.21% w/wExplanation:
Given:
Mass of solute = 30 gram
Volume of water = 500 ml
Density = 1.15g/ml
Find:
(i)ppm
(ii) %w/v
(iii) %w/w
Computation:
Water in gram = 500 ml × 1.15 g/ml
Water in gram = 575 gram
In ppm
Concentration of MgSO4 = [30 / 575] × 10⁶
Concentration of MgSO4 = 0.0521 × 10⁶ ppm
in % w/v
Concentration of MgSO4 = [30 / 500] × 100
Concentration of MgSO4 = 6% w/v
in % w/w
Concentration of MgSO4 = [30 / 575] × 100
Concentration of MgSO4 = 5.21% w/w
whats the ph for a solution poh4 9.78 concentration of solution
Answer:
4.22
Explanation:
According to the question, the pOH of the solution is 9.78. You may recall that pOH is the hydroxide concentration of a solution.
Also pOH = -log[OH^-]. Hence the pOH is obtained from the hydroxide ion concentration.
Finally, pH + pOH =14
Hence;
pH = 14-pOH
pH= 14-9.78 = 4.22
pH= 4.22
please help guys the question is
give reasons
a. we have to separate the mixture
b. All impure substances are not harmful.
c. A mixture of iron fillings and sand can be separated by using a magnet
d. A sentences "shake before well use" is written on the bottle of the medicine.
Answer:
(a )people separate mixtures in order to ger a specific substance that they need.
Arrange the compounds in order of decreasing magnitude of lattice energy:
a. LiBr
b. KI
c. CaO.
Rank from largest to smallest.
Answer:
The correct answer is CaO > LiBr > KI.
Explanation:
Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.
With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.
The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.
Arranging the chemical compounds in order of decreasing magnitude of lattice energy, we have:
c. CaO.
a. LiBr
b. KI
Lattice energy can be defined as a measure of the energy required to dissociate one (1) mole of an ionic compound into its constituent anions and cations, in the gaseous state.
Hence, it is typically used to measure the bond strength of ionic compounds.
Generally, lattice energy is inversely proportional to the size of the ions and directly proportional to their electric charges.
Lithium bromide (LiBr) comprises the following ions:
[tex]Li^+[/tex] and [tex]Br^-[/tex]Potassium iodide (KI) comprises the following ions:
[tex]K^+[/tex] and [tex]I^-[/tex]Calcium oxide (CaO) comprises the following ions:
[tex]Ca^{2+}[/tex] and [tex]O^{2-}[/tex]From the above, we can deduce that there is an increase in the charge possessed by the ionic chemical compounds and as such this would result in an increase in the lattice energy.
In order of decreasing magnitude of lattice energy, the chemical compounds are arranged as:
I. CaO.
II. KI.
III. LiBr.
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For element radon, give the chemical symbol, atomic number, and group number.
Which of the following chemical equations corresponds to the standard molar enthalpy of formation of Na_2CO_3(s)?
a. 2 NA(s) + C(s) + 3 O(g) ------------> Na_2CO_3(s)
b. Na_2O(s) + CO_2(g) --------------->Na_2CO_3 (s)
c. Na_2(s) + C(s) + 3 O(g) -------------> Na_2CO_3 (s)
d. Na_2O(s) + CO(g) ---------------> Na_2CO_3(s)
e. 2 Na(s)+ C(s) + 3/2 O_2(g) ------------> Na_2CO_3(s)
Answer:
2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)
Explanation:
The molar enthalpy of formation of a chemical is defined as the change in enthalpy during the formation of 1 mole of the substance from its constituent elements (Constituent elements are pure elements you have in the periodic table)
For Na₂CO₃ constituent elements are Na(s), C(s) and O₂(g) and the chemical equation that represents the molar enthalpy is:
2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)Match the tools with the advantages they offer to astronomers.
photography
space telescope
radio telescope
optical telescope
w
detects electromagnetic frequencies outside
the visible spectrum that reaches Earth
captures images that can be shared and
compared by scientists
obtains a magnified and clear view of a part
of the sky to observe celestial objects
allows access to images taken from outside
Earth's atmosphere
Answer:
- photography (captures images that can be shared and
compared by scientists)
- space telescope (allows access to images taken from outside
Earth's atmosphere)
- radio telescope (detects electromagnetic frequencies outside
the visible spectrum that reaches Earth)
- optical telescope (obtains a magnified and clear view of a part
of the sky to observe celestial objects)
Explanation:
Photography is used to capture still images based on the principle that some compounds react in the presence of optical energy.
Space telescope is a type of observatory telescope positioned in outer space to observe distant planets, galaxies and other astronomical objects. Space telescopes reduces the interference from ultraviolet frequencies, X-rays and gamma rays; as well as light pollution which ground-based observatories encounter.
A radio telescope is a specialized antenna and radio receiver used to receive radio waves from astronomical radio sources in the sky. Radio telescope is used to study radio frequencies emitted by astronomical objects, that fall outside the visible light spectrum.
An optical telescope is used to gather and focuses light, from a far distant object. Optical telescope is used within the visible light spectrum of the electromagnetic spectrum, to create a magnified image, for direct view, or to make a photograph, or to collect data through electronic image sensors.
Answer:
- photography (captures images that can be shared and
compared by scientists)
- space telescope (allows access to images taken from outside
Earth's atmosphere)
- radio telescope (detects electromagnetic frequencies outside
the visible spectrum that reaches Earth)
- optical telescope (obtains a magnified and clear view of a part
of the sky to observe celestial objects)
Explanation:
Assume that you are provided with the following materials:
• Strips of metallic zinc, metallic copper, metallic iron
• 1M aqueous solutions of ZnSO4, CuSO4, FeSO4, and aqueous iodine(I2)
• Other required materials to create Voltaic cells such as beakers, porous containers, graphite rods, a voltmeter, and a few wires with alligator clips.
In this modified version of the lab, after thoroughly studying the lab hand out and watching the videos,identify 4 different combinations of Voltaic cells that are possible to be created with the above materials.For each cell created, include the following details.
A) Which electrode was the anode,and which was the Cathode?
B) The anode and cathode half reactions.
C) Balanced equation for each cell you propose to construct.
D) Calculated Eocelle Short hand notation (line notation) for each cell (be sure to include the inactive electrode if needed).
Answer:
See explanation
Explanation:
First voltaic cell;
Zn(s)|Zn^2+(aq)||Cu^2+(aq)|Cu(s)
Anode;
Zinc
Cathode;
Copper
Oxidation half equation;
Zn(s)------> Zn^2+(aq) + 2e
Reduction half equation;
Cu^2+(aq) +2e -----> Cu(s)
Overall; Zn(s) + Cu^2+(aq) -----> Zn^2+(aq) + Cu(s)
E°cell = 0.34 -(-0.76) =1.1 V
Second voltaic cell;
Zn(s)|Zn^2+(aq)||Fe^2+(aq)|Fe(s)
Anode;
Zinc
Cathode;
Iron
Oxidation half equation;
Zn(s)------> Zn^2+(aq) + 2e
Reduction half equation;
Fe^2+(aq) +2e -----> Fe(s)
Overall; Zn(s) + Fe^2+(aq) -----> Zn^2+(aq) + Fe(s)
E°cell = (-0.44) -(-0.76) = 0.32 V
Third voltaic cell;
Fe(s)|Fe^2+(aq)||Cu^2+(aq)|Cu(s)
Anode;
Iron
Cathode;
Copper
Oxidation half equation;
Fe(s)------> Fe^2+(aq) + 2e
Reduction half equation;
Cu^2+(aq) +2e -----> Cu(s)
Overall; Fe(s) + Cu^2+(aq) -----> Fe^2+(aq) + Cu(s)
E°cell = 0.34 -(-0.44) = 0.78 V
Fourth voltaic cell
Cu(s)|Cu^2+(aq)||I2(aq)|C(s)|I^-(aq)
Anode;
Copper
Cathode;
Graphite rod
Oxidation half equation;
Cu(s)------> Cu^2+(aq) + 2e
Reduction half equation;
I2(aq) +2e -----> 2I^-(aq)
Overall; Cu(s) + I2(aq) -----> Cu^2+(aq) + 2I^-(aq)
E°cell = 0.54 -0.34 = 0.20 V
What is the rate constant of a reaction if rate = 1 x 10-2 (mol/L)/s, [A] is 2 M,
[B] is 3 M, m = 2, and n = 1?
Answer:
[tex]0.10 \text{ L$^2$mol$^{-2}$s$^{-1}$}[/tex]
Explanation:
The general formula for a rate law is
[tex]\text{rate} = k\text{[A]}^m \text{[B]}^{n}[/tex]
With your numbers, the rate law becomes
1.2 mol·L⁻¹s⁻¹ = k(2 mol·L⁻¹)²(3 mol·L⁻¹)¹ = k × 4 mol²L⁻² × 3 mol·L⁻¹
= 12k mol³L⁻³
[tex]\\ k = \dfrac{\text{1.2 mol $\cdot$ L$^{-1}$s$^{-1}$} }{12\text{ mol$^{3}$L}^{-3}} = \mathbf{0.10} \textbf{ L$\mathbf{^2}$mol$^{\mathbf{-2}}$s$^{\mathbf{-1}}$}[/tex]
Enter your answer in the provided box.
The equilibrium constant KP for the reaction
CO(g) + Cl2(g) ⇌ COCl2(g)
is 5.62 × 1035 at 25°C. Calculate ΔG
o
f
for COCl2 at 25°C.
Answer:
The correct answer is -341.2 kJ per mole.
Explanation:
The reaction given is:
CO (g) + Cl₂ (g) ⇔ COCl₂ (g)
Kp = 5.62 × 10³⁵
T = 25 °C or 298 K
The formula for calculating ΔG is,
ΔG° = -RTlnKp
ΔG° = -8.314 × 298 ln (5.62 × 10^35)
ΔG° = -203.9 kJ/mol
ΔG° = ∑nΔG°f (products) -∑nΔG°f (reactants)
ΔG° = ΔG°f (COCl₂ (g)) - [ΔG°f (CO(g)) + ΔG°f (Cl₂(g))]
ΔG°f (COCl₂ (g)) = ΔG° + [ΔG°f (CO (g)) + ΔG°f (Cl₂(g))]
ΔG°f (COCl₂ (g)) = -203.9 + (-137.28 + 0.00)
ΔG°f (COCl₂ (g)) = -341.2 kJ/mol
The standard Gibbs free energy [tex]\mathbf{\Delta G^o_f}[/tex] for COCl2 at 25°C is -341.25 kJ/mol
The given equation for the chemical reaction is
CO(g) + Cl2(g) ⇌ COCl2(g)
At the temperature of 25°C = (273 + 25) K, the equilibrium constant [tex]\mathbf{K_p = 5.62\times 10^{35}}[/tex]
Consider the expression for the relationship between [tex]\mathbf{\Delta G^o}[/tex] and [tex]\mathbf{K_p }[/tex] for the equilibrium reaction can be expressed as:
[tex]\mathbf{\Delta G^o = - RT In K_p}[/tex]
where;
gas constant (R) = 8.314 × 10⁻³ kJ/K.mol∴
[tex]\mathbf{\Delta G^o = - (8.314 \times 10^{-3}\ kJ/K.mol \times 298 \ K) \times In (5.62 \times 10^{35} )}[/tex]
[tex]\mathbf{\Delta G^o = -2.477572\ K \times 82.31680992}[/tex]
[tex]\mathbf{\Delta G^o = 203.95 \ kJ}[/tex]
Thus, the standard free energy for the reaction is 203.95 kJ/mol
For a given reaction, the standard Gibbs free energy can be calculated by using the formula:
[tex]\mathbf{\Delta G^o_{rxn} = \sum n \Delta G^o_f (products) - \sum m \Delta G^o_f (reactants) }[/tex]
[tex]\mathbf{\Delta G^o_{rxn} =\Big [\Delta G^o_{f} (COCl_{2(g)} ) -\Big(\Delta G^o_{f} (CO)_{(g)} + \Delta G^o_{f} (Cl)_{2(g)} ) \Big ) \Big ] }[/tex]
replacing the values of and solving for COCl2 at standard free energy of formation of substances, we have:
[tex]\mathbf{-203.95 \ kJ/mol =\Big [\Delta G^o_{f} (COCl_{2(g)} ) -\Big(-137.3 kJ/mol + 0 \ kJ/mol\Big ) \Big ] }[/tex]
Collecting like terms, we have:
[tex]\mathbf{\Delta G^o_{f} (COCl_{2(g)} ) = -203.95 \ kJ/mol -137.3 kJ/mol }[/tex]
[tex]\mathbf{\Delta G^o_{f} (COCl_{2(g)} ) = -341.25 \ kJ/mol }[/tex]
Therefore, we can conclude that the standard Gibbs free energy [tex]\mathbf{\Delta G^o_f}[/tex] for COCl2 at 25°C is -341.25 kJ/mol
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(9443+45−9.9) (9443+45−9.9) ×8.4× 10 6
For the following reaction, 61.6 grams of bromine are allowed to react with 25.5 grams of chlorine gas. bromine (g) + chlorine (g) bromine monochloride (g) What is the maximum amount of bromine monochloride that can be formed? grams
Answer:
I need great answers
Explanation:
please rate my answer as great
Which is an intensive property of a substance?
Answer:
length
Explanation:
edge 2020
hope this helps!
Answer:
A.) Density
Explanation:
Correct on edge.
A galvanic cell consists of a Cu(s)|Cu2+(aq) half-cell and a Cd(s)|Cd2+(aq) half-cell connected by a salt bridge. Oxidation occurs in the cadmium half-cell. The cell can be represented in standard notation as
Answer:
[tex]Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)[/tex]
Explanation:
A galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs.
The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.
As it is given that cadmium acts as anode, it must be on the left hand side and copper must be on right hand side.
[tex]Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)[/tex]
If the Ksp for Li3PO4 is 5.9×10−17, and the lithium ion concentration in solution is 0.0020 M, what does the phosphate concentration need to be for a precipitate to occur?
Answer:
7.4 × 10⁻⁹ M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for Li₃PO₄: 5.9 × 10⁻¹⁷
Concentration of lithium ion: 0.0020 M
Step 2: Write the reaction for the solution of Li₃PO₄
Li₃PO₄(s) ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
Step 3: Calculate the phosphate concentration required for a precipitate to occur
The solubility product constant is:
Ksp = 5.9 × 10⁻¹⁷ = [Li⁺]³ × [PO₄³⁻]
[PO₄³⁻] = 5.9 × 10⁻¹⁷ / [Li⁺]³
[PO₄³⁻] = 5.9 × 10⁻¹⁷ / 0.0020³
[PO₄³⁻] = 7.4 × 10⁻⁹ M
How many moles of gaseous boron trifluoride, BF3, are contained in a 4.3421 L bulb at 787.9 K if the pressure is 1.218 atm?
Answer:
The amount of moles of gaseous boron trifluoride, BF₃, contained in a 4.3421 L bulb at 787.9 K if the pressure is 1,218 atm is 0.082 moles
Explanation:
An ideal gas is a theoretical gas that is considered to be made up of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P*V = n*R*T
where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.
In this case:
P= 1.218 atmV= 4.3421 Ln= ?R= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]T= 787.9 KReplacing:
1.218 atm* 4.3421 L= n*0.082 [tex]\frac{atm*L}{mol*K}[/tex] *787.9 K
Solving:
[tex]n=\frac{1.218 atm* 4.3421 L}{0.082 \frac{atm*L}{mol*K}*787.9 K}[/tex]
n= 0.082 moles
The amount of moles of gaseous boron trifluoride, BF₃, contained in a 4.3421 L bulb at 787.9 K if the pressure is 1,218 atm is 0.082 moles
A 30.5 g sample of a compound contains 9.29 g of nitrogen and the rest is oxygen. What is the empirical formula of the compound?
Answer:
The empirical formula of the compound is NO2.
Explanation:
The following data were obtained from the question:
Mass of compound = 30.5 g
Mass of nitrogen (N) = 9.29 g
Empirical formula of compound =?
Next, we shall determine the mass of oxygen in the compound. This can be obtained as follow:
Mass of compound = 30.5 g
Mass of nitrogen (N) = 9.29 g
Mass of oxygen (O) =?
Mass of O = mass of compound – mass of N.
Mass of O = 30.5 – 9.29
Mass of O = 21.21 g
Finally, we shall determine the empirical formula of the compound as follow:
Mass of nitrogen (N) = 9.29 g
Mass of oxygen (O) = 21.21 g
Divide by their molar mass.
N = 9.29 / 14 = 0.664
O = 21.21 / 16 = 1.326
Divide by the smallest
N = 0.664/ 0.664 = 1
O = 1.326/ 0.664 = 2
Therefore the empirical formula of the compound is NO2.
A sample of ice absorbs 15.6kJ of heat as it undergoes a reversible phase transition to form liquid water at 0∘C. What is the entropy change for this process in units of JK? Report your answer to three significant figures. Use −273.15∘C for absolute zero.
Answer:
Entropy change of ice changing to water at 0°C is equal to 57.1 J/K
Explanation:
When a substance undergoes a phase change, it occurs at constant temperature.
The entropy change Δs, is given by the formula below;
Δs = q/T
where q is the quantity of heat absorbed or evolved in Joules and T is temperature in Kelvin at which the phase change occur
From the given data, T = 0°C = 273.15 K, q = 15.6 KJ = 15600 J
Δs = 15600 J / 273.15 K
Δs = 57.111 J/K
Therefore, entropy change of ice changing to water at 0°C is equal to 57.1 J/K
The entropy change of ice changing to water will be "57.1 J/K".
Entropy changeThe shift in what seems like a thermodynamic system's condition of confusion is caused by the transformation of heat as well as enthalpy towards activity. Entropy seems to be greater mostly in a network with a high quantity or measure of chaos.
According to the question,
Temperature, T = 0°C or,
= 273.15 K
Heat, q = 15.6 KJ or,
= 15600 J
We know the formula,
Entropy change, Δs = [tex]\frac{q}{T}[/tex]
By substituting the values, we get
= [tex]\frac{15600}{273.15}[/tex]
= 57.11 J/K
Thus the above answer is correct.
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Akeem cut his finger during an investigation, and it is bleeding slightly. Before helping him bandage the wound,
which precaution should the teacher take?
O Tell someone to call 911,
O Put on protective gloves.
O Wash Akeem's finger in the shower.
O Apply disinfectant before cleaning.
Answer:
b.) Put on protective gloves
Answer:
2020 Put on protective gloves.
Explanation:
How many mL of a 0.130 M aqueous solution of chromium(II) nitrate, Cr(NO3)2, must be taken to obtain 5.08 grams of the salt
Answer:
222.3 ml of a 0.130 M aqueous solution of chromium (II) nitrate must be taken to obtain 5.08 grams of the salt.
Explanation:
Being:
Cr: 52 g/moleN: 14 g/moleO: 16 g/molethe molar mass of chromium (II) nitrate, Cr(NO₃)₂ is:
Cr(NO₃)₂ = 52 g/mole + 2* (14 g/mole + 3* 16 g/mole)= 176 g/mole
So: if 176 grams are present in 1 mole of the compound, 5.08 grams in how many moles of the compound will be present?
[tex]amount of moles=\frac{5.08 grams* 1 mole}{176 grams}[/tex]
amount of moles=0.0289 moles
Molarity (M) is the number of moles of solute that are dissolved in a given volume. It is then calculated by dividing the moles of the solute by the volume of the solution:
[tex]molarity (M)=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in [tex]\frac{moles}{liter}[/tex]
So in this case:
molarity= 0.130 Mnumber of moles of solute= 0.0289 molesvolume= ?Replacing:
[tex]0.130 M=0.130 \frac{moles}{liter} =\frac{0.0289 moles}{volume}[/tex]
Solving:
[tex]volume=\frac{0.0289 moles}{0.130 \frac{moles}{liter} }[/tex]
volume=0.2223 liters
Being 1 L= 1,000 mL:
volume=0.222 liters= 222.3 mL
222.3 ml of a 0.130 M aqueous solution of chromium (II) nitrate must be taken to obtain 5.08 grams of the salt.
A student determines the value of the equilibrium constant to be 1.5297 x 107 for the following reaction: HBr(g) + 1/2 Cl2(g) --> HCl(g) +1/2 Br2(g) Based on this value of Keq, calculate the Gibbs free energy change for the reaction of 2.37 moles of HBr(g) at standard conditions at 298 K.
Answer:
[tex]\Delta G=-97.14kJ[/tex]
Explanation:
Hello,
In this case, the relationship between the equilibrium constant and the Gibbs free energy of reaction is:
[tex]\Delta G=-RTln(K)[/tex]
Hence, we compute it as required:
[tex]\Delta G=-8.314\frac{J}{mol\times K}*298K*ln(1.5297x10^7)\\\\\Delta G=-40.99kJ/mol[/tex]
And for 2.37 moles of hydrogen bromide, we obtain:
[tex]\Delta G=-40.99kJ/mol*2.37mol\\\\\Delta G=-97.14kJ[/tex]
Best regards.
Which of the following combinations will result in a reaction that is spontaneous at all temperatures?
Negative enthalpy change and negative entropy change
Negative enthalpy change and positive entropy change
Positive enthalpy change and negative entropy change
Positive enthalpy change and positive entropy change
PLS EXPLAIN WHAT EACH MEANS AND THE VARIABLES AND THE EXPLANATION BEHIND IT
Answer:
[tex]\huge\boxed{Option \ 2}[/tex]
Explanation:
A reaction is spontaneous at all temperatures by the following combinations:
=> A negative enthalpy change ( [tex]\triangle H < 0[/tex] )
=> A positive entropy change ( [tex]\triangle S > 0[/tex] )
See the attached file for more better understanding!
from Gibbs Equation, [tex] \Delta G = \Delta H - T\Delta S [/tex]
reaction is spontaneous if $\Delta G$ is negative.
so, first option is not valid at high temperature, ($-h+ts$)
second, is always a spontaneous reaction, ($-h-ts$)
third, is never spontaneous ($+h+ts$)
4th is similar to second, spontaneous at higher temperatures ($+h-ts$)
Match each term to an appropriate example. (4 points) Column A 1. Gravitational energy : Gravitational energy 2. Nuclear energy : Nuclear energy 3. Radiant energy : Radiant energy 4. Stored mechanical energy : Stored mechanical energy Column B a. Fission and fusion b. A ball at the top of a hill c. A compressed spring d. X-rays and light
Answer:
Explanation:
Before proceeding to answering the questions, let's try and understand each term.
Gravitational energy can be defined as the potential energy an object has when placed in a high position (compared to been placed in a low position). Examples include; a knife been placed at the top of the cupboard, a ball at the top of a hill.
Nuclear energy can be defined as the energy released or consumed as a result of splitting or joining together of the nuclei (plural of nucleus) of atom(s). The process of splitting an atomic nucleus is called nuclear fission while the process of joining two nuclei of atoms is called nuclear fusion.
Radiant energy can be defined as the energy of electromagnetic wave. One of the properties of electromagnetic waves is that they travel through space. Examples include X-rays and light
Stored mechanical energy can be defined as the potential energy an object has as a result of the application of force. Examples include; a stretched rubber band, a compressed spring
What happens if we put raw eggs in a pot full of hot oil?
A sailor on a trans-Pacific solo voyage notices one day that if he puts 735.mL of fresh water into a plastic cup weighing 25.0g, the cup floats in the seawater around his boat with the fresh water inside the cup at exactly the same level as the seawater outside the cup (see sketch at right).
Calculate the amount of salt dissolved in each liter of seawater. Be sure your answer has a unit symbol, if needed, and round it to 2 significant digits.
You'll need to know that the density of fresh water at the temperature of the sea around the sailor is 0.999/gcm3. You'll also want to remember Archimedes' Principle, that objects float when they displace a mass of water equal to their own mass.
Answer:
Amount of salt in 1 L seawater = 34 g
Explanation:
According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater
mass of freshwater = density * volume
1 cm³ = 1 mL
mass of freshwater = 0.999 g/cm³ * 735 cm³ = 734.265 g
mass of freshwater + cup = 734.265 + 25 = 759.265 g
Therefore, mass of equal volume of seawater = 759.265 g
Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)
1 liter = 1000 cm³ = 1000 mL;
Density of seawater = mass / volume
Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L
Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L
mass of 1 Liter seawater = 1033.01 g
mass of 1 Liter freshwater = 999 g
mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g
Therefore, amount of salt in 1 L seawater = 34 g
Identify the compound that does NOT have hydrogen bonding.
A) CH3NH2
B) H2O
C) (CH3)3N
D) CH3OH
E) HF
Answer:
(CH3)3N
Explanation:
Hydrogen bonding can be called a type of intracellular force of the attraction. It is the force that occur between molecules. It is the bonding between the molecules and of hydrogen and electronegative items in the covalent bond. This is called the hydrogen donor. An electro-negative hydrogen atoms may be a hydrogen bonded. It is also called a hydrogen acceptor.
Thus in (CH3)3N, the hydrogen atoms becomes bonded with carbon. Carbon is not electronegative atoms. Thus it does not play as donor. Nitrogen is electronegative and play as hydrogen acceptor. But there is no presence of hydrogen acceptor. Thus there is no molecules that exhibit hydrogen molecules bonding.
[tex]\bold {(CH_3)_3N}[/tex] does not have hydrogen bonding because of the absence of electronegativity difference.
Hydrogen bond:
It is an inter-molecular bond. It is due to the difference in electronegativities of constituent atoms. This creates dipole in the atoms so, atoms start to attract each other.
In [tex]\bold {(CH_3)_3N}[/tex], the hydrogen atoms are bonded with carbon. The difference between the electronegativities Carbon and hydrogen is very less.
Therefore, [tex]\bold {(CH_3)_3N}[/tex] does not have hydrogen bonding because of the absence of electronegativity difference.
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A solution contains 2.2 × 10-3 M in Cu2+ and 0.33 M in LiCN. If the Kf for Cu(CN)42- is 1.0 × 1025, how much copper ion remains at equilibrium?
Answer:
[Cu²⁺] = 2.01x10⁻²⁶
Explanation:
The equilibrium of Cu(CN)₄²⁻ is:
Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻
And Kf is defined as:
Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴
As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:
[Cu²⁺] = 0
[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M
[Cu(CN)₄²⁻] = 2.2x10⁻³
Some [Cu²⁺] will be formed and equilibrium concentrations will be:
[Cu²⁺] = X
[CN⁻] = 0.3212M + 4X
[Cu(CN)₄²⁻] = 2.2x10⁻³ - X
Where X is reaction coordinate
Replacing in Kf equation:
1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴
1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0
Solving for X:
X = 2.01x10⁻²⁶
As
[Cu²⁺] = X
[Cu²⁺] = 2.01x10⁻²⁶Suppose a student completes an experiment with an average value of 2.9 mL and a calculated standard deviation of 0.71 mL. What is the minimum value within a 1 SD range of the average
Answer:
The correct answer is 2.2 mL.
Explanation:
Given:
Average: 2.9 mL
SD: 0.71 mL
We can define a 1 SD range in which the value of volume (in mL) will be comprised:
Volume (mL) = Average ± SD = (2.9 ± 0.7) mL
Maximum value= Average + SD= 2.9 + 0.7 mL = 3.6 mL
Minimum value= Average - SD = 2.9 - 0.7 mL = 2.2 mL
Thus, the minimum value within a 1 SD range of the average is 2.2 mL
The minimum value within 1 SD is 2.19 mL
The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x\ is\ raw\ score, \mu=mean,\sigma=standard\ deviation[/tex]
Given that μ = 2.9 mL, σ = 0.71 mL; hence:
The minimum value within 1 SD range = μ ± σ = 2.9 ± 0.71 = (2.19, 3.61)
Therefore the minimum value within 1 SD is 2.19 mL
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A piece of solid Fe metal is put into an aqueous solution of Cu(NO3)2. Write the net ionic equation for any single-replacement redox reaction that may be predicted. Assume that the oxidation state of in the resulted solution is 2 . (Use the lowest possible coefficients for the reaction. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank. If no reaction occurs, leave all boxes blank and click on Submit.)
Answer:
Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)
Explanation:
The ionic equation shows the actual reaction that took place. It excludes the spectator ions. Spectator ions are ions that do not really participate in the reaction even though they are present in the system.
For the reaction between iron and copper II nitrate, the molecular reaction equation is;
Fe(s) + Cu(NO3)2(aq)----> Fe(NO3)2(aq) +Cu(s)
Ionically;
Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)
2. Find the two generic molecules from Part 1 that are made of 3 atoms. a. Compare and contrast these two molecules by listing two similarities and two differences.
Answer:
hello the molecules are missing from your question below are the Generic molecules : [tex]ABE_{3}[/tex] and [tex]AB_{3} E[/tex]
answer : It can be determined that both generic molecules are polar
It can be determined that both generic molecules have similar molecular shape
They have different Geometry
They differ in bond angles as well
Explanation:
The two generic molecules : [tex]ABE_{3}[/tex] and [tex]AB_{3} E[/tex]
comparing(similarities) these two generic molecules
It can be determined that both generic molecules are polar
It can be determined that both generic molecules have similar molecular shape
differences between the generic molecules
They have different Geometry
They differ in bond angles as well
Increasing which factor will cause the gravitational force between two objects to decrease?
weights of the objects
distance between the objects
acceleration of the objects
masses of the objects
Answer:
B
Explanation:
Increasing distance between the objects factor will cause the gravitational force between two objects to decrease. Therefore, option B is correct.
What causes gravitational force to decrease?The gravitational force grows in proportion to the size of the masses . The gravitational force weakens rapidly as the distance between masses grows. Unless at least one of the objects has a lot of mass, detecting gravitational force is extremely difficult.
Gravity is affected by object size and distance between objects. Mass is a unit of measurement for the amount of matter in an object.
The force of gravity is proportional to the masses of the two objects and inversely proportional to the square of the distance between them. This means that the force of gravity increases with mass but decreases as the distance between objects increases.
Thus, option B is correct.
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