Answer:
1.30 m
Explanation:
First we calculate the number of moles of solute in the solution, using the definition of molarity:
Molarity = moles / litersConverting 250 mL ⇒ 250 / 1000 = 0.250 Lmoles = 1.50 M * 0.250 L = 0.375 molesThen we calculate the mass of solvent, using the given volume and density:
1.15 g/mL * 250 mL = 287.5 gConverting 287.5 g to kg ⇒ 287.5 / 1000 = 0.2875 kgNow we calculate the molality of the solution:
molality = moles of solute / kg of solvent = 0.375 mol / 0.2875 kgmolality = 1.30 mBased on the equations below, which metal is the least active? Pb(NO3)2(aq) + Ni (s) --> Ni(NO3)2 (aq)+ Pb(s) Pb(NO3)2(aq) + Ag(s) --> No reaction Cu(
Answer:
Ni
Explanation:
An active metal is a highly reactive metal. Active metals are found high up in the activity series.
Active metals react with other metals that are lower than them in the activity thereby displacing the lower metals from a solution of their salts. This is what may have happened in the other two reactions.
Ni is the most active metal listed in the question since it can react a compounds with Pb(NO3)2(aq) to liberate Pb metal.
The mass of a single tantalum atom is 3.01×10-22 grams. How many tantalum atoms would there be in 37.1 milligrams of tantalum?
Answer: There are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.
Explanation:
Given: Mass of single tantalum atom = [tex]3.01 \times 10^{-22} g[/tex]
Mass of tantalum atoms = 37.1 mg (1 mg = 0.001 g) = 0.0371 g
Therefore, number of tantalum atoms present in 0.0371 grams is calculated as follows.
[tex]No. of atoms = \frac{0.0371 g}{3.01 \times 10^{-22}}\\= 1.23 \times 10^{20}[/tex]
Thus, we can conclude that there are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.
There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.
Explanation:
Given:
Mass of single atom of tantalum =[tex]3.01\times 10^{-22} g[/tex]
To find:
The number of atoms of tantalum in 37.1 milligrams.
Solution:
Mass of tantalum = 37.1 mg
[tex]1 mg = 0.001 g\\37.1 mg=37.1\times 0.001 g=0.0371 g[/tex]
The number of atoms in 0.0371 grams of tantalum = N
Mass of a single atom of tantalum = [tex]3.01\times 10^{-22} g[/tex]
Then a mass of N atoms of tantalum will be:
[tex]0.0371 g=N\times 3.01\times 10^{-22} g\\N=\frac{0.0371 g}{ 3.01\times 10^{-22} g}\\=1.23\times 10^{20}[/tex]
There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.
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In a closed system, If a gas is transported to a container with double the volume of the previous container, the gas was held in, what is the gases' new volume?
The volume of the gas is fixed and will not change.
The volume of the gas will be half the original volume.
The volume of the gas will be the original volume squared.
The volume of the gas will be double the original volume.
Answer:
The volume of the gas is fixed and will not change.
Explanation:
The volume of the gas will not change because there is no change in temperature. Temperature increases the volume of gases enclosed in a container.
One main difference between the heating of gases on the one hand and solids or liquids on the other is that ___________________. One main difference between the heating of gases on the one hand and solids or liquids on the other is that ___________________. heating of gases depends not only on the temperature difference, but also on the process as well as the amount of gas present. heating of gases depends on temperature difference as well as the amount of gas present. specific heat is not defined for gases. heat cannot be exchanged with gases.
Answer:
heating of gases depends not only on the temperature difference, but also on the process as well as the amount of gas present.
Explanation:
The work done when a gas is heated does not only depends on the initial and final states of the gas but also on the process used to achieve the change of state of the gas.
Several processes can be applied in changing the state of a gas such as; adiabatic process, isobaric process, isochoric process and isothermal process.
Hence, the heating of a gas, depends not only on the temperature difference, as well as the amount of gas present according to the ideal gas laws but also on the process used to achieve the change of state.
Na2CO3 reacts with dil.HCl to produce NaCl, H2O and CO2. If 21.2 g of pure Na2CO3 are added in a solution containing 21.9g HCl , a. Find the limiting reagent. (2) b. Calculate the number of moles of excess reagent left over.(2) c. Calculate the number of molecules of H2O formed.(1) d. Calculate volume of CO2 gas produced at 270C and 760mm Hg pressure.(2) e. Write significance of limiting reagent
Answer:
See explanation
Explanation:
Equation of the reaction;
Na2CO3(aq) + 2HCl(aq) -------> 2NaCl(aq) + H2O(l) + CO2(g)
Number of moles of Na2CO3 = 21.2g/106g/mol = 0.2 moles Na2CO3
Number of moles of HCl = 21.9g/36.5g/mol = 0.6 moles of HCl
1 mole of Na2CO3 reacts with 2 moles of HCl
0.2 moles of Na2CO3 reacts with 0.2 × 2/1 = 0.4 moles of HCl
Hence Na2CO3 is the limiting reactant
Since there is 0.6 moles of HCl present, the number of moles of excess reagent=
0.6 moles - 0.4 moles = 0.2 moles of HCl
1 mole of Na2CO3 forms 1 mole of water
0.2 moles of Na2CO3 forms 0.2 moles of water
Number of molecules of water formed = 0.2 moles × 6.02 × 10^23 = 1.2 × 10^23 molecules of water
1 mole of Na2CO3 yields 1 mole of CO2
0.2 moles of Na2CO3 yields 0.2 moles of CO2
1 mole of CO2 occupies 22.4 L
0.2 moles of CO2 occupies 0.2 × 22.4 = 4.48 L at STP
Hence;
V1=4.48 L
T1 = 273 K
P1= 760 mmHg
T2 = 27°C + 273 = 300 K
P2 = 760 mmHg
V2 =
P1V1/T1 = P2V2/T2
P1V1T2 = P2V2T1
V2 = P1V1T2/P2T1
V2 = 760 × 4.48 × 300/760 × 273
V2= 4.9 L
The limiting reactant is the reactant that determines the amount of product formed in a reaction. When the limiting reactant is exhausted, the reaction stops.
Methyl orange can change color by transitioning from one chromophore to another. When added to a clear solution and the solution turns red, it is determined to be a(n) __________ in its __________ stable form.
Answer:
acidic titration in its stable form
Explanation:
Methyl orange can change its color in titration solution. The yellow color is towards alkaline solution and red color is towards acidic solution. The Ph value of solution will change during this chemical process.
The data shows the number of years that 30 employees worked for an insurance company before retirement. is the population mean for the number of years worked, and % of the employees worked for the company for at least 10 years. (Round off your answers to the nearest integer.)
Answer:
14
73%
Explanation:
The mean Number of years worked :
. (sum of service years) / employees in the
(8+13+15+3+13+28+4+12+4+26+29+3+10+3+17+13+15+15+23+13+12+1+14+14+17+16+7+27+18+24) /
(417 / 30)
= 13.9 years
= 14 years
The percentage of employees who have worked for atleast 10 years :
Number of employees with service years ≥ 10 years = 22 employees
Total number of employees
Percentage (%) = (22 / 30= * 100% = 0.7333 * 100% = 73.33% = 73%
A student dropped a piece of nickel metal into a solution of HCl(aq). He observed the formation of gas bubbles and collected the gas into another test tube. The student performed a splint test and observed that the splint flared up when he placed the splint into the test tube of the gas. What can be said about the results of this students experiment?
a. The student performed the splint test incorrectly. He should of observed a popping sound when the splint was placed in the test tube.
b. The experiment was performed incorrectly. Nickel doesn't react with HCl. Therefore, the student picked up the wrong metal when conducting the experiment.
c. The student completed the experiment correctly and there were no errors in the experiment.
d. The student performed the splint test incorrectly. He should of observed the flame being extinguished when the splint was placed in the test tube.
Answer:
a. The student performed the splint test incorrectly. He should of observed a popping sound when the splint was placed in the test tube.
Explanation:
It is given that a student performed an experiment where he dropped a nickel metal in to HCl solution. He observed the reaction and performed a splint test in the test tube that is filled with a gas which is formed while Nickle is dropped into the solution of HCl.
But the experiment that the student performed was incorrect. He must have observed the popping sound when the splint was placed in the test tube.
When the splint was added to the gas splint flared up. The hydrogen gas pops out when exposed to the flame.
[tex]$Ni + HCl(aq) = NiCl + H_2$[/tex]
Thus the correct option is (a).
1 or 2 topics or two lessons should be explained in an illustrated childrens book minimum of 10 pages must have 3 or more sentences
Answer:
Yes because same topic are long
The doctor has ordered Claforan 1 g in 100 ml D5W to run IV piggyback for 30 minutes twice daily. The pharmacy sends Claforn 2 g in a powdered form, which when reconstituted has a concentration of 180 mg Claforan per ml. How much Claforn will you add to the bag of D5W
Answer:
0.111 g
Explanation:
1 g = 1000 mg
Doctor ordered the following concentration of Claforan:
C = 1 g/100 mL x 1000 mg/1 g = 10 mg/mL
If we add 2 g iof Claforan, we obtain:
2 g Claforn ---- 180 mg/mL Claforan
To reach a concentration equal to C (10 mg/mL), we need:
10 mg/mL Claforan x 2 g Claforn/(180 mg/mL Claforan) = 0.111 g Claforn
Therefore, we have to add 0,111 g (111 mg) of Claforn to the bag of 100 ml D5W to obtain the ordered concentration of 10 mg/mL Claforan.
Classify each of the following as either macroscopic, microscopic or particulate:
a. a red blood cell.
b. a sugar molecule.
c. baking powder.
Answer:
Classify each of the following as either macroscopic, microscopic or particulate:
a. a red blood cell.
b. a sugar molecule.
c. baking powder.
Explanation:
a. A red blood cell is a microscopic particle.
It can be viewed under a microscope.
b. A sugar molecule is also a microscopic substance.
It can be viewed under a microscope.
c. Baking powder is macroscopic substance.
Calculate the average atomic mass for X
Answer:
39.0229 amu
Explanation:
Hello there!
In this case, according to given information, the idea here is to multiply the percent abundance by the mass number of each isotope and then add them all together as shown below:
[tex]=0.0967*38+0.7868*39+0.1134*40+0.0031*41\\\\=3.6746+30.6852+4.536+0.1271\\\\=39.0229amu[/tex]
Regards!
Perform the following operation and express the answer in scientific notation.
7.296×10² ÷ 9.6×10^-9
Answer:
7.6×10¹⁰
Explanation:
7.296×10²÷9.6×10⁻⁹
To solve such problem,
We group the whole number ans solved seperately and also group the indices and solve the seperately
Step1 : 7.296/9.6 = 0.76
Step 2: applying the law of indices,
10²÷10⁻⁹ = 10⁽²⁺⁹⁾ = 10¹¹
Therefore,
7.296×10²÷9.6×10⁻⁹ = 0.76×10¹¹ = 7.6×10¹⁰
What does the term spontaneous mean in chemical reactions?
A. Producing heat as a product
B. Occurring without added energy
C. Occurring only at high temperatures
D. Occurring in an aqueous solution
Answer:
B
Explanation:
Spontaneous in chemical reactions means without any external input.
Occurring without added energy. Hence, option B is correct.
What is a spontaneous reaction?A spontaneous reaction is a reaction that supports the formation of products under the conditions under which the reaction is happening.
Spontaneous Reaction- a reaction that favours the formation of products at the conditions under which the reaction is occurring.
A non-spontaneous reaction can be made spontaneous if it is inside a controlled environment, this is what happens in nuclear power plants that create atomic fusion and fission in chambers that are controlled to control different particles to create nuclear active rays.
Hence, option B is correct.
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How many molecules in each sample?
64.7 g N2
83 g CCl4
19 g C6H12O6
Answer:
1.39x10²⁴ molecules N₂.25x10²³ molecules CCl₄6.38x10²² molecules C₆H₁₂O₆Explanation:
First we convert the given masses into moles, using the compounds' respective molar mass:
64.7 g N₂ ÷ 28 g/mol = 2.31 mol N₂83 g CCl₄ ÷ 153.82 g/mol = 0.540 mol CCl₄19 g C₆H₁₂O₆ ÷ 180 g/mol = 0.106 mol C₆H₁₂O₆Then we multiply each amount by Avogadro's number, to calculate the number of molecules:
2.31 mol N₂ * 6.023x10²³ molecules/mol = 1.39x10²⁴ molecules0.540 mol CCl₄ * 6.023x10²³ molecules/mol = 3.25x10²³ molecules0.106 mol C₆H₁₂O₆ * 6.023x10²³ molecules/mol = 6.38x10²² molecules14 protons,14 electrons and 14 neutrons
the answer is silicon!!
When electrons in a molecule are not found between a pair of atoms but move throughout the molecule, this is called Group of answer choices
Answer:
delocalised electrons
Explanation:
they are called delocalised electrons because that can move freely in the molecule
I have an unknown volume of gas held at a temperature of 115 K in a container with a pressure of 60atm. If by increasing the temperature to 225 K and decreasing the pressure to 30. atm causes the volume of the gas to be 29 liters, how many liters of gas did I start with?
SHOW YOUR WORK
Explanation:
here is the answer to your question.
A
(c) 2 C(s) + MnO2(s)
Mn(s) + 2 CO(g)
O combination reaction
O decomposition reaction
O combustion reaction
O single-displacement reaction
Answer: The reaction, [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex] is a single-displacement reaction.
Explanation:
A chemical reaction in which one element of a compound is replaced by another element participating in the reaction.
For example, [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex]
Here, the element manganese is replaced by carbon atom. As only one element gets replaced so, it is a single-displacement reaction.
Thus, we can conclude that [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex] is a single-displacement reaction.
Question 16(Multiple Choice Worth 5 points)
(04.01 LC) Which statement is true about the total mass of the reactants during a chemical change?
O It is destroyed during chemical reaction.
O It is less than the total mass of the products. O It is equal to the total mass of the products.
O It is greater than the total mass of the products.
Answer:
It is equal to the total mass of the products.
Explanation:
Hope this helps :)
Water has a density of 1.00 g/mL. If you put an object that has a density of 0.79 g/mL into water, it will sink to the
bottom.
ANSWER:
True
False
Answer:
False
Explanation:
An object with a density greater than 1.00g/mL (greater than the density of water) will sink. An object with a density less than the density of water, will float.
If the water has a density of 1.00 g/mL. If you put an object that has a density of 0.79 g/mL into water, it will sink to the bottom, this statement is false.
What is density?The density of an actual content is its mass per unit volume. The most common symbol for density is d, but the Latin letter D can also be used.
Three of an object's most fundamental properties are mass, volume, and density. Mass describes how heavy something is, volume describes its size, and density is defined as mass divided by volume.
The density of something is a measure of how heavy it is in relation to its size. When an artifact is more dense than water, it plunges; when an object is less dense than water, it floats.
Density is a property of a substance that is independent of the amount of substance.
As in the given scenario, water is having density 1 g/mL and object in having density less then it so it will float on water.
Thus, the given statement is false as the material will not sink, rather it will float on water.
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a sample of copper was heated at 275.1 C and placed into 272 g of water at 21.0 C. The temperature of the water rose at 29.7 C. How many grams of copper were in the sample
Answer:
104.8 g
Explanation:
From the question given above, the following data were obtained:
Initial temperature of copper (T꜀) = 275.1 °C
Mass of water (Mᵥᵥ) = 272 g
Initial temperature of water (Tᵥᵥ) = 21 °C
Equilibrium temperature (Tₑ) = 29.7 °C
Mass of copper (M꜀) =?
NOTE:
Specific heat capacity of copper (C꜀) = 0.385 J/gºC
Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC
Finally, we shall determine the mass of the copper in the sample. This can be obtained as follow:
Heat loss by copper = Heat gained by water
M꜀C꜀(T꜀ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
M꜀ × 0.385 (275.1 – 29.7) = 272 × 4.184(29.7 – 21)
M꜀ × 0.385 × 245.4 = 1138.048 × 8.7
M꜀ × 94.479 = 9901.0176
Divide both side by 94.479
M꜀ = 9901.0176 / 94.479
M꜀ = 104.8 g
Thus, the mass of the copper in the sample is 104.8 g
what is the mass of insoluble calcium phosphate produced from .555 grams of calcium chloride
Answer:
0.518 g
Explanation:
Step 1: Write the balanced equation
3 CaCl₂ + 2 H₃PO₄ ⇒ Ca₃(PO₄)₂ + 6 HCl
Step 2: Calculate the moles corresponding to 0.555 g of CaCl₂
The molar mass of CaCl₂ is 110.98 g/mol.
0.555 g × 1 mol/110.98 g = 5.00 × 10⁻³ mol
Step 3: Calculate the moles of Ca₃(PO₄)₂ produced
5.00 × 10⁻³ mol CaCl₂ × 1 mol Ca₃(PO₄)₂/3 mol CaCl₂ = 1.67 × 10⁻³ mol Ca₃(PO₄)₂
Step 4: Calculate the mass corresponding to 1.67 × 10⁻³ moles of Ca₃(PO₄)₂
The molar mass of Ca₃(PO₄)₂ is 310.18 g/mol.
1.67 × 10⁻³ mol × 310.18 g/mol = 0.518 g
The pKa of an acid can be determined through _____ (reduction, titration, filtration) with a strong base.
Gradually increase the volume of the base, stopping _____ (before, as, after) the equivalence point is reached.
The pKa of the acid is equal to the pH at the _____ (equivalence point, midway volume to the equivalence point, maximum volume).
Answer:
The pKa of an acid can be determined through titration with a strong base.
Gradually increase the volume of the base, stopping before the equivalence point is reached.
The pKa of the acid is equal to the pH at the midway volume to the equivalence point.
Explanation:
An acid HA dissociates in water as follows:
HA ⇄ H⁺ + A⁻ Ka
So, it produces hydrogen ions (H⁺) and a conjugate base (A⁻). The concentrations of HA, H⁺ and A⁻ at equilibrium determine the constant Ka. The pKa is calculated as:
pKa = -log Ka
The relationship between the pH of the solution and the pKa of the acid is described by the Henderson-Hasselbalch equation:
pH = pKa + log ([A⁻]/[HA])
The pKa can be experimentally determined by acid-base titration, in which a strong base is added to the acid solution. As the base is added, the acid HA is neutralized and the conjugate base A⁻ is formed. Thus, the concentration of the acid ([HA]) increases and the concentration of the conjugate base ([A⁻] decreases. The equivalence point is reached when the total amount of acid is neutralized with the added base. Before reaching the equivalence point, at the halfway point, half of the acid is neutralized and converted into the conjugate base. Thus:
[A-] = [HA] ⇒ log [A-]/[HA] = log 1 = 0 ⇒ pH = pKa
We measure the pH at that point and it is equal to the pKa of the acid.
Match the description with the type of precipitation being described.
1. Its formation requires very strong updrafts
2. Its formation requires falling through a layer of above freezing air
3. Precipitation from cumuliform clouds is typically of this nature
4. Precipitation from stratus clouds is typically of this nature
Options:
a. Hail
b. Drizzle
c. Shower
d. Freezing Rain
Answer:
1. Its formation requires very strong updrafts = a. Hail
2. Its formation requires falling through a layer of above-freezing air = d. Freezing Rain
3. Precipitation from cumuliform clouds is typically of this nature = c. Shower
4. Precipitation from stratus clouds is typically of this nature = Drizzle
Explanation:
Hail formation requires very strong updrafts, these updrafts are the upward moving air created in a thunderstorm. This period of noticeable thunderstorms creates hails.
Freezing rain requires the presence of warm air, it requires falling through a layer of above-freezing air to the colder air below to produce an ice coating on anything it drops on.
Showers are produced by cumuliform clouds which look like cotton balls. Since cumuliform clouds precipitate too, these clouds can have fluctuating rain in a day in the form of showers.
Drizzle which raises low visibility is considered a type of liquid precipitation since it also falls from a cloud. Drizzle which is obviously smaller in diameter when compared to that of raindrops, however, is common with stratus clouds.
A buffer is prepared containing 0.75 M NH3 and 0.20 M NH4 . Calculate the pH of the buffer using the Kb for NH3. g
Answer:
pH=8.676
Explanation:
Given:
0.75 M [tex]NH_{3}[/tex]
0.20 M [tex]NH_{4}[/tex]
The objective is to calculate the pH of the buffer using the kb for [tex]NH_3[/tex]
Formula used:
[tex]pOH=pka+log\frac{[salt]}{[base]}\\[/tex]
pH=14-pOH
Solution:
On substituting salt=0.75 and base=0.20 in the formula
[tex]pOH=-log(1.77*10^{-5})+log\frac{0.75}{0.20}\\ =4.75+0.5740\\ =5.324[/tex]
pH=14-pOH
On substituting the pOH value in the above expression,
pH=14-5.324
Therefore,
pH=8.676
Calculate the moment of inertia of a CH³⁵CL₃ molecule around a rotational axis that contains the C-H bond. The C-Cl bond length is 177pm and the HCCl angle is 107⁰f
Answer:
The correct answer is "[tex]4.991\times 10^{-45} \ kg.m^2[/tex]".
Explanation:
According to the question,
[tex]R_{C-Cl} = 177 \ pm[/tex]
or,
[tex]=1.77\times 10^{-10} \ m[/tex]
[tex]\alpha = 107^{\circ}[/tex]
[tex]m_{Cl}=34.97 \ m.u[/tex]
or,
[tex]=34.97\times 1.66\times 10^{-27}[/tex]
[tex]=5.807\times 10^{-26} \ kg[/tex]
The moment of inertia around the rotational axis will be:
⇒ [tex]I=3\times m_{Cl}\times (R_{C-Cl})^2 \ Sin^2 \alpha[/tex]
By putting the values, we get
[tex]=3\times 5.807\times 10^{-26}\times (1.77\times 10^{-10})^2 \ Sin^2 (107)[/tex]
[tex]=3\times 5.807\times 10^{-26}\times (1.77\times 10^{-10})^2\times 0.91452[/tex]
[tex]=4.991\times 10^{-45} \ kg.m^2[/tex]
Please help me ASAP I’ll mark Brainly
Answer:
1. Vacuole
2. chloroplast
3. Nucleus
4. Plasma membrane - cell membrane
5. Vacuole (same as #1 ?) could be vesicle
Explanation:
What is different between margerine and butter in term of organic chemistry
Answer:
The most important difference between the two is that butter is derived from dairy and is rich in saturated fats, whereas margarine is made from plant oils. ... If the margarine contains partially hydrogenated oils, it will contain trans fat, even if the label claims that it has 0 g.
Explanation:
(⌒_⌒;)
According to the Arrhenius equation, changing which factors will affect the
rate constant?
A. Temperature and the ideal gas constant
B. The activation energy and the constant A
C. The constant A and the temperature
D. Temperature and activation energy
Answer:
e−(Ea/RT): the fraction of the molecules present in a gas which have energies equal to or in excess of activation energy at a particular temperature
Answer:
D. Temperature and activation energy is the correct answer
Explanation:
^_^
