Answer:
Filtration is a technique used as a remedy to separate mixes
Explanation:
If you have a sodium, then you will explore that salt dissolves but the sand is still the same.
If the salt in the resin water solution scanners, the sand remains the residue and passes through the filter paper.
All you have to do now is pleasant the salty water so that the water can evaporate, leaving the salt behind.
In lipid bilayers, there is an order-to-disorder transition similar to the melting of a crystal. Comparing a lipid bilayer that is made up mostly of palmitoleic acid to one made up mostly of palmitic acid, which would have a higher melting transition temperature
Answer:
The lipid bilayer made up of Palmitic acid will have a higher melting transition temperature
Explanation:
The one with a higher melting transition temperature is the lipid layer with a higher melting temperature
Melting temperature of palmitoleic acid = -0.5°C
Melting temperature of palmitic acid = 62.9°C
Hence the lipid bilayer made up of Palmitic acid will have a higher melting transition temperature
Under certain conditions, the substance mercury(II) oxide can be broken down to form mercury and oxygen. If 32.2 grams of mercury(II) oxide react to form 29.8 grams of mercury, how many grams of oxygen must simultaneously be formed
Explanation:
This is a decomposition reaction. Firstly, you will want to write the chemical equation out and balance it.
[tex]2Hg_2O->4Hg+O_2[/tex] (The -> is supposed to be an arrow, sorry!)
We see that there's only 1mol of Oxygen made in the products, we can do some simple math to solve for the amount of grams of Oxygen produced according to the amount of the reactant (Hg2O).
[tex]32.2gHg_2O*\frac{1molHg_2O}{417.18gHg_2O}*\frac{1molO_2}{2molHg_2O}*\frac{32gO_2}{1molO_2}[/tex]
I want to break this down, just in case:
The 417.18gHg2O is the molecular mass of the molecule (so I doubled Hg and added 16 to it to get this number).
As we can see in the chemical equation, 1mol Hg2O produces 2mol O because Oxygen is a diatomic molecule (so there will always be two of it when it's by itself).
And finally, in 1mol O2 there are 32g of O2.
** When you do math like this, always make sure that all of your units cancel out except for the units you're looking for. For example, here we're looking for the grams of Oxygen, so after everything else cancels out, we should only have grams O2.
So, 1.23gO2 should be your answer.
Name the following aldehyde PLEASE PLEASE HELP
Answer:
Explanation:
Answer is D 2,5-dimethylheptanal
You should accern the lowest possible number close to the parent name
For each of the following circumstances, indicate whether the calculated molarity of NaOH would be lower, higher or unaffected. Explain your answer in each case. a. The inside of the pipet used to transfer the standard HCl solution was wet with water.b. you added 40 mL of water to the titration flask rather than 25ml. c. The buret, wet with water, was not rinsed with NaOH solution before filling the buret with NaOH solution. d. Five (5) drops of phenolphthalein were added to the solution to be titrated rather than three (3) drops.
Answer:
a)calculated molarity of NaOH would be lower
b) calculated molarity of NaOH would be lower
c) calculated molarity of NaOH would be lower
d) calculated molarity of NaOH would be unaffected
Explanation:
Let us recall that the reaction of NaOH and HCl is as follows;
NaOH(aq) + HCl(aq) ----> NaCl(aq) + H2O(l)
Since the reaction is 1:1, when the number of moles of HCl reacting with NaOH is low due to dilution, the calculated molarity of NaOH also becomes less than it's accurate value.
When 40mL of water is added to the titration flask rather than 25ml of water, the acid is more dilute hence less number of moles of acid than necessary reacts with the base thereby yielding a less than accurate value of the molarity of NaOH.
If the burette wet with water is not rinsed with NaOH solution, the concentration of the NaOH in the burette decreases due to dilution with water and a less than accuracy value is calculated for the molarity of NaOH.
If five drops of phenolphthalein is used instead of one or two drops, there is no qualms since enough phenolphthalein may be added to ensure that a sharp end point is obtained.
Rank the following compounds in order of decreasing boiling point: sodium chloride (NaCl), methane (CH4), and iodomethane (CH3I). Rank from highest to lowest boiling point.
Answer:
CH4< CH4I< NaCl
Explanation
NaCl has the boiling point of 1,413°C ( 2,575°F )
CH3I has a boiling point of 42°C ( 107°F )
CH4 has the boiling point of -161.6°C ( -258.9°F )
why might the melting point of the crystals obtained in this experiment be close to but below one of the reference melting points and melt slowly over several degrees
Answer: hello the experiment related to your question is missing but I will provide a more general answer within the scope of your question
answer :
presence of Impurities
Explanation:
The melting point of the crystals as obtained in the experiment will be close to but below reference melting points and will also melt slower because of the presence of impurities in the compound
Impurities alter the melting and freezing points from ideal freezing and melting points of compounds
How many miles are in 8.73 *10^25 atoms of boron
The correct question is: How many moles are in [tex]8.73 \times 10^{25}[/tex] atoms of boron.
Answer: There are 145 moles present in [tex]8.73 \times 10^{25}[/tex] atoms of boron.
Explanation:
According to the mole concept, there are [tex]6.022 \times 10^{23}[/tex] atoms present in one mole of every substance.
Hence, number of moles present in [tex]8.73 \times 10^{25}[/tex] atoms is calculated as follows.
[tex]Moles = \frac{8.73 \times 10^{25}}{6.022 \times 10^{23}}\\= 1.45 \times 10^{2}\\= 145 mol[/tex]
Thus, we can conclude that there are 145 moles present in [tex]8.73 \times 10^{25}[/tex] atoms of boron.
To identify a halide, you can react a solution with chlorine water in the presence of mineral oil.
If the unknown halide is a
Choose...weaker/stronger
reducing agent than chlorine, the halide will be oxidized to
Choose...its elemental form/ its ionic form/ a solid
which would change the color of the
Choose...mineral oil/ aqueous
layer.
Answer:
- Stronger reducing agent than Chlorine
- Oxidized to it's elemental form
- Change the colour of the aqueous layer.
Explanation:
Halides are electronegative elements in group seven of the Periodic table which have gained electrons to complete their electronic configuration.
They include F-, CL-, Br- and I-.
As you descend the group electro negativity decreases as the number of outermost shells increases. Hence F- is the most electronegative while I- is the least electronegative.
In terms of oxidising and reducing abilities amongst the halogens, since an oxidizing agent readily accepts electrons and is thereby reduced, oxidizing power decreases down the group.
For example, Fluorine being the strongest oxidising agent in the group readily accepts electrons from other members of the group and is reduced to the fluoride ion
F + e = F -
Therefore in terms of oxidizing abilities,
F > Cl > Br > I
Conversely, , as the oxidising power decreases down the group, the reducing powers increases
Therefore, in terms of reducing powers,
I > Br > Cl > F
In the test for halide ions using aqueous chlorine, since chlorine is a stronger oxidizing agent/weaker reducing agent than Bromine or iodine, it readily accepts their electrons forming the chloride ion.
Cl2 + 2 Br- = 2 Cl- + Br2
The bromide ion (assuming the unknown halide is bromide) being a stronger reducing agent/weaker oxidizing agent than Chlorine would readily lose it's electrons and get oxidized to it's elemental form changing the colour of the aqueous layer to brown.
That is : Br2- = Br2 + 2e
The fill in the blanks could be filled with stronger, elemental form and mineral oil.
Identification of a halide:In the case when the halide is not known so it should be stronger. The halide should be oxidized with respect to the elemental form and it should change the color of mineral oil. Due to this, halide should be oxidized for elemental halogen i.e. more soluble for mineral oil.
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The majority of metals are found within
The majority of metals are found in ores.
But a few such as copper, gold, platinum, and silver frequently occur in the free state because they do not readily react with other elements.
A. Directions:Describe ways on how to protect humans and plants and animals Choose your answer in the box and fill in the concept map below
Answer:
i) Humans - drinks lot of water
- use umbrella
- use hats and sunglasses
- apply sunblock when swimming
- wear rush guard
ii) Plants - water the plants
-put a shade on the plants
iii) Animals - provide pets with plenty of water
- place the animals under the tree during noon time
En un recipiente cerrado y rígido se introdujo una mezcla gaseosa a cierta temperatura y las presiones parciales de cada gas son: p(F2) = 2,00 atm, p(BrF) = 1,50 atm y p(BrF3) = 0,0150 atm. A la temperatura que se preparó la mezcla tiene lugar la reacción representada por:
BrF3 (g) ⇌ BrF(g) + F2(g) Kp(T) = 64,0
Elegir la afirmación correcta.
Seleccione una:
Qp > Kp, por lo tanto, las presiones parciales de BrF(g) y F2(g) aumentan hasta alcanzar el equilibrio.
Qp < Kp, por lo tanto, la presión parcial de BrF3(g) disminuye hasta alcanzar el equilibrio.
Qp = Kp, por lo tanto, las presiones parciales de BrF3(g), BrF(g) y F2(g) no cambian.
Qp < Kp, por lo tanto, las presiones parciales de BrF(g) y F2(g) disminuyen hasta alcanzar el equilibrio.
Qp > Kp, por lo tanto, la presión parcial de BrF3(g) aumenta hasta alcanzar el equilibrio.
Answer:
Qp > Kp, por lo tanto, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.
Explanation:
Paso 1: Escribir la ecuación balanceada
BrF₃ (g) ⇌ BrF(g) + F₂(g) Kp(T) = 64,0
Paso 2: Calcular el cociente de reacción (Qp)
Qp = pBrF × pF₂ / pBrF₃
Qp = 1,50 × 2,00 / 0,0150 = 200
Paso 3: Sacar una conclusión
Dado que Qp > Kp, la reacción se desplazará hacia la izquierda para alcanzar el equilibrio, es decir, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.
Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . Suppose 35. g of hydrochloric acid is mixed with 73.0 g of sodium hydroxide. Calculate the minimum mass of hydrochloric acid that could be left over by the chemical reaction. Round your answer to significant digits.
Answer:
No mass of HCl could be left over by the chemical reaction because is the limting reactant and it is all consumed.
Explanation:
Our reactants are: HCl and NaOH
Products are: NaCl and H₂O
This is a neutralization reaction that can also be called an acid base reaction, an acid and a base react to produce water and a neutral salt, in this case where we have strong acid and base.
Ratio is 1:1. We convert mass to moles:
35 g . 1 mol / 36.45 g = 0.960 moles of HCl
73 g . 1 mol / 40 g = 1.82 moles of NaOH
As ratio is 1:1, for 0.960 moles of HCl we need 0.960 moles of NaOH and for 1.82 moles of NaOH, we need 1.82 moles of acid.
As we only have 0.960 moles of HCl and we need 1.82 moles, no acid remains after the reaction goes complete. HCl is the limiting reactant, so the acid, it is all consumed.
What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:
2 NaN3 is) 2Na (s) + 3N2 (8)
Answer:
108.6 g
Explanation:
2NaN₃(s) → 2Na(s) + 3N₂(g)First we use the PV=nRT formula to calculate the number of nitrogen moles:
P = 1.00 atmV = 56.0 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 0 °C ⇒ 0 + 273.2 = 273.2 KInputting the data:
1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 Kn = 2.5 molThen we convert 2.5 moles of N₂ into moles of NaN₃, using the stoichiometric coefficients of the balanced reaction:
2.5 mol N₂ * [tex]\frac{2molNaN_3}{3molN_2}[/tex] = 1.67 mol NaN₃Finally we convert 1.67 moles of NaN₃ into grams, using its molar mass:
1.67 mol * 65 g/mol = 108.6 gWhat is the largest number
(of a single species) a specific area can support?
A. Population
B. Unlimited factor
C. Carrying capacity
D. Niche
Answer:
carrying capacity
Explanation:
Thus, the carrying capacity is the maximum number of individuals of a species that an environment can support. Population size decreases above carrying capacity due to a range of factors depending on the species concerned, but can include insufficient space, food supply, or sunlight.
A comparison analysis subjects a suspect specimen and a control specimen to the same tests and examinations for the ultimate purpose of determining: whether or not they have a common origin if they are identical in chemical composition if the same person handled them if they are alike in molecular structure All of these
Answer:
whether or not they have a common origin
Explanation:
During the scientific investigation of a crime, a forensic scientist might be required to perform a comparison analysis. A comparison analysis is an examination where specific physical properties are obtained from a suspect specimen and a standard which are then compared to identify their common origin. This type of analysis is important for it helps in narrowing down the crime to a particular person. It also enables accurate delivery of justice.
How many moles of carbon dioxide at Stp will fit in a 50 liter container?
Answer:
If the gas is at STP, THE 1 mole is 22.4 liters.
Explanation:
Consider the following chemical equilibrium: Now write an equation below that shows how to calculate from for this reaction at an absolute temperature . You can assume is comfortably above room temperature. If you include any common physical constants in your equation be sure you use their standard symbols, found in the ALEKS Calculator.
Answer:
Kp=Kc *(RT)+-3
Explanation:
The relation between Kp and Kc is given below:
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant , R = 0.082057 L atm.mol⁻¹K⁻¹
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
Δn = (0)-(2+1) = -3
Thus, Kp is:
Kp=Kc *(RT)+-3
determine the mass in grams of 3.75 x 10^21 atoms of zinc. (the mass of one mole of zinc is 65.39 g)
Answer: The mass in [tex]3.75 \times 10^{21}[/tex] atoms of zinc is 0.405 g.
Explanation:
Given: Atoms of zinc = [tex]3.75 \times 10^{21}[/tex]
It is known that 1 mole of every substance contains [tex]6.022 \times 10^{23}[/tex] atoms. So, the number of moles in given number of atoms is as follows.
[tex]Moles = \frac{3.75 \times 10^{21}}{6.022 \times 10^{23}}\\= 0.622 \times 10^{-2}\\= 0.0062 mol[/tex]
As moles is the mass of a substance divided by its molar mass. So, mass of zinc (molar mass = 65.39 g/mol) is calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\0.0062 mol = \frac{mass}{65.39 g}\\mass = 0.405 g[/tex]
Thus, we can conclude that the mass in [tex]3.75 \times 10^{21}[/tex] atoms of zinc is 0.405 g.
From the dropdowns, identify whether the compound contains ionic bonds, covalent bonds, or both. a) CBr4 [ Select ] b) copper(II) sulfate [ Select ] c) N2O3 [ Select ] d) phosphorous trichloride
Answer:
a) Covalent bonds
b) Covalent and ionic bonds
c) Covalent bonds
d) Covalent bonds
Explanation:
Metals and non-metals form ionic bonds (electrons are transferred), whereas nonmetals and nonmetals form covalent bonds.
Identify whether the compound contains ionic bonds, covalent bonds, or both.
a) CBr₄. C and Br are nonmetals. Thus, they form covalent bonds.
b) copper(II) sulfate. Sulfate contains S and O (nonmetals), which are bonded through covalent bonds. Sulfate is bonded to copper (metal) through an ionic bond.
c) N₂O₃. N and O are nonmetals. Thus, they form covalent bonds.
d) phosphorous trichloride. P and Cl are nonmetals. Thus, they form covalent bonds.
An aqueous solution contains 0.29 M of benzoic acid (HA) and 0.16 M of sodium benzoate (A-). If the pH of this solution was measured to be 4.63, calculate the pKa of benzoic acid g
Answer:
pKa = 4.89.
Explanation:
We can solve this problem by using the Henderson-Hasselbach equation, which states:
pH = pKa + log [tex]\frac{[A^-]}{[HA]}[/tex]
In this case [A⁻] is the concentration of sodium benzoate and [HA] is the concentration of benzoic acid.
We input the given data:
4.63 = pKa + log [tex]\frac{0.16}{0.29}[/tex]
And solve for pKa:
pKa = 4.89
17. The density of a population would influence which limiting factor?
O niche
O growth rate
O weather
O space
Answer:
The answer is growth rate
Explanation:
it will help you
Calculate the mass of Na2S needed if a solution containing 2g of Hg(NO3)2 was added to Na2S solution.
( Hg= 200.59, N= 14, O= 16, Na= 23, S=32)
Answer:
1.433g of HgS are produced
Explanation:
A Solution Containing 2.0 Grams Of Hg(NO3)2 Was Added To A Solution Containing 2.0 Grams Of Na2S. Calculate the mass of the HgS that was formed (it is a precipitate) according to this reaction:
Based on the reaction:
Na2S + Hg(NO3)2 → HgS + 2NaNO3
To solve this question we need to find the moles of each reactant in order to find the limiting reactant. The moles of limiting reactant = moles of HgS:
Moles Na2S -Molar mass: 78.0452 g/mol-
2.0g * (1mol / 78.0452g) = 0.0256 moles Na2S
Moles Hg(NO3)2 -324.7g/mol-
2.0g * (1mol / 324.7g) = 0.006159 moles Hg(NO3)2
As the reaction is 1:1, and moles of Hg(NO3)2 < moles Na2S
The moles of Hg(NO3)2 = Moles HgS = 0.006159 moles
The mass is:
Mass HgS -Molar mass: 232.66g/mol-:
0.006159 moles * (232.66g/mol) =
1.433g of HgS are produced
chemistry help!
This mysterious gas has a volume of 4.35 L and a pressure of 1.20 atm. If the pressure is changed to 0.95 atm, what will the new volume be? How do i set up this problem and solve it?
Refer to the attachment.
Hope this helps you...
Potassium carbonate, K 2CO 3, sodium iodide, NaI, potassium bromide, KBr, methanol, CH 3OH, and ammonium chloride, NH 4Cl, are soluble in water. Which produces the largest number of dissolved particles per mole of dissolved solute
Answer:
Potassium carbonate (K₂CO₃)
Explanation:
The compounds dissociate into ions in water, as follows:
K₂CO₃ → 2 K⁺ + CO₃⁻ ⇒ 3 dissolved particles per mole
NaI → Na⁺ + I⁻ ⇒ 2 dissolved particles per mole
KBr → K⁺ + Br⁻ ⇒ 2 dissolved particles per mole
CH₃OH → CH₃O⁻ + H⁺ ⇒ 2 dissolved particles per mole
NH₄Cl → NH₄⁺ + Cl⁻ ⇒ 2 dissolved particles per mole
Therefore, the largest number of dissolved particles per mole of dissolved solute is produced by potassium carbonate (K₂CO₃).
PLEASE HELP QUICKLY
The diagram shows the potential energy changes for a reaction pathway. (10 points)
Part 1: Does the diagram illustrate an endothermic or an exothermic reaction? Give reasons in support of your answer.
Part 2: Describe how you can determine the total change in enthalpy and activation energy from the diagram and if each is positive or negative.
The activation energy of the reaction is the difference between the highest point on the reaction profile and the energy of the reactants.
What is a potential energy diagram?A potential energy diagram or a reaction profile shows us the energ change between the reactants and the products.
As we look at the reaction profile, we observe that the products have a greater energy than the reactants hence the reaction is endothermic. The enthalpy chamgeis obtained by subtracting the energy of the products from the energy of the reactants.
The activation energy of the reaction is the difference between the highest point on the reaction profile and the energy of the reactants.
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A student dropped a pea size amount of K2CO3 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 was extingished 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 completed the experiment correctly and there were no errors in the experiment.
b. The experiment was performed incorrectly. K2CO3 doesn't react with HCl. Therefore, the student picked up the wrong compound when conducting the experiment.
c. The student performed the splint test incorrectly. He should of observed the splint flare up when the splint was placed in the test tube.
d. The student performed the splint test incorrectly. He should of observed a popping sound when the splint was placed in the test tube.
Answer:
The student completed the experiment correctly and there were no errors in the experiment.
Explanation:
When a pea size amount of K2CO3 is dropped into a solution of HCl, the following reaction occurs;
K2CO3(s) + 2HCl(aq) ----> 2KCl(aq) + CO2(g) + H2O(l)
The gas CO2 does not support burning hence, when the student performed a splint test and observed that the splint was extinguished when he placed the splint into the test tube of the gas.
Hence, the experiment was properly conducted and the student completed the experiment correctly and there were no errors in the experiment.
Write two balanced half-equations for this redox equation:
2Cu + O2 = 2CuO
Answer:
2 Cu + O2 → 2 CuO
This is an oxidation-reduction (redox) reaction:
2 Cu0 - 4 e- → 2 CuII
(oxidation)
2 O0 + 4 e- → 2 O-II
(reduction)
Cu is a reducing agent, O2 is an oxidizing agent.
A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was , calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.
Answer:
molar heat of combustion = -5156 *10³ kJ/mol
Explanation:
A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.
Step 1: Data given
Mass of naphthalene = 1.435 grams
Initial temperature of water = 20.28 °C
Final temperature of water = 25.95 °C
heat capacity of the bomb plus water was 10.17 kJ/°C
Molar mass naphtalene = 128.2 g/mol
Step 2:
Qcal = Ccal * ΔT
⇒with Qcal =the heat of combustion
⇒with Ccal = heat capacity of the bomb plus water = 10.17 kJ/°C
⇒with ΔT = the difference in temperature = T2 - T1 = 25.95 - 20.28 = 5.67°C
Qcal = 10.17 kJ/°C * 5.67 °C
Qcal = 57.7 kJ
Step 3: Calculate moles
Moles naphthalene = 1.435 grams / 128.2 g/mol
Moles naphthalene = 0.01119 moles
Step 4: Calculate the molar heat of combustion
molar heat of combustion = Qcal/ moles
molar heat of combustion = -57.7 kJ/ 0.01119 moles
molar heat of combustion = -5156 *10³ kJ/mol
Give your familiarity for following terms
1. roasting 2. smelting 4. zone refining 5. polling
Answer:
The roasting process is a delicate combination of art and science . Roasters are familiar with how the beans look and the smells Well, familiarity with the machine makes things much easier to predict, but the best way is to do many different tests .Well, familiarity with the machine makes things much easier to predict, but the best way is to do many different tests.Smelting is a process of applying heat to ore in order to extract a base metal. It is a form of extractive metallurgy. It is used to extract many metals from their ores, including silver, iron, copper, and other base metals.In zone refining, solutes are segregated at one end of the ingot in order to purify the remainder, or to concentrate the impurities. For example, in the preparation of a transistor or diode semiconductor, an ingot of germanium is first purified by zone refining. In zone refining, solutes are segregated at one end of the ingot in order to purify the remainder, or to concentrate the impurities. ... For example, in the preparation of a transistor or diode semiconductor, an ingot of germanium is first purified by zone refining.Polling is the process where the computer or controlling device waits for an external device to check for its readiness or state, often with low-level hardware. For example, when a printer is connected via a parallel port, the computer waits until the printer has received the next character.Explanation:
hope it heloed
Qual número atômico do ferro?
Answer:
The answer is Iron.
Explanation:
I hope this helps you out. Have a nice day!
