You need to prepare a solution with a specific concentration of Na+Na+ ions; however, someone used the end of the stock solution of NaClNaCl, and there isn’t any NaClNaCl to be found in the lab. You do, however, have some Na2SO4Na2SO4. Can you substitute the same number of grams of Na2SO4Na2SO4 for the NaClNaCl in a solution? Why or why not?

Answers

Answer 1

Answer:

Explanation:

Ionic equation

NaCl(aq) --> Na+(aq) + Cl-(aq)

Na2SO4(aq) --> 2Na+(aq) + SO4^2-(aq)

In NaCl solution, 1 mole of Na+ is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na+ is dissociated in 1 liter of solution.

Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)

= 142 g/mol

Molecular weight of NaCl = 23 + 35.5

= 58.5 g/mol

Masses

% Mass of NA+ in Na2SO4 = mass of Na+/total mass of Na2SO4 * 100

= 46/142 * 100

= 32.4%

% Mass of NA+ in NaCl = mass of Na+/total mass of NaCl * 100

= 23/58.5 * 100

= 39.3%

Therefore, the % mass of Na+ in NaCl and Na2SO4 are different so it cannot be used.

Answer 2

Answer:

Final answer:

You cannot substitute Na2SO4 directly for NaCl based on mass since they have different molar masses. The same mass of Na2SO4 will provide more Na+ ions than NaCl, leading to a change in the Na+ ion concentration.

Explanation:

No, you cannot substitute the same number of grams of Na2SO4 for the NaCl in a solution. This is because NaCl and Na2SO4 have different molar masses and therefore different numbers of moles per gram. The concentration of a solution is determined by the number of moles of solute per unit volume of solvent, not the mass. Hence, using the same mass of a different compound would alter the concentration of Na+ ions in the solution.

For instance, if one mole of NaCl gives us one mole of Na+, one mole of Na2SO4 will provide two moles of Na+. In other words, the same mass of Na2SO4 contains more Na+ ions than the same mass of NaCl. So using the same mass of Na2SO4 in place of NaCl will result in a solution with a higher Na+ ion concentration.

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Mg(OH)2 is a sparingly soluble salt with a solubility product, Ksp, of 5.61×10−11. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams. What is the ratio of solubility of Mg(OH)2 dissolved in pure H2O to Mg(OH)2 dissolved in a 0.180 M NaOH solution? Express your answer numerically as the ratio of molar solubility in H2O to the molar solubility in NaOH.

Answers

Answer:

The ration of the molar solubility is 165068.49.

Explanation:

The solubility reaction of the magnesium hydroxide in the pure water is as follows.

$Mg(OH)_(2)\Leftrightarrow Mg^(2+)(aq)+2(OH)^(-)(aq)$

$[Mg^(2+)][OH^(-)]$

Initial 0 0

Equili +S +2S

Final S 2S

$K_(sp)=[Mg^(2+)][OH^(-)]$

$5.61* 10^(-11)=(S)(2S)^(2)$

$S=((5.61* 10^(-11))/(4))^(1/3)=2.41* 10^(-4)M$

Solubility of $Mg(OH)_(2)$ in 0.180 M NaOH is a follows.

$Mg(OH)_(2)\Leftrightarrow Mg^(2+)(aq)+2(OH)^(-)(aq)$

$[Mg^(2+)][OH^(-)]$

Initial 0 0

Equili +S +2S

Final S 2S+0.180M

$K_(sp)=[Mg^(2+)][OH^(-)]$

$5.61* 10^(-11)=(S)(2S+0.180)^(2)$

$S=1.46* 10^(-9)M$

$Ratio\,of\,solubility=(2.41* 10^(-4))/(1.46* 10^(-9))=165068.49$

Therefore, The ration of the molar solubility is 165068.49.

What action is NOT necessary before lighting a Bunsen burner?A)Make sure there are no holes in the hose connected to the burner.
B)Leave the gas stop open for a few minutes before lighting to clear the line.
C)Clear the burner area of flammable materials such as solvents and papers.
D)Check that the gas regulator and air inlet on the burner can be adjusted.

Answers

The Bunsen burner is a heating apparatus used to heat samples in the laboratory. The correct answer is option B: )Leave the gas stop open for a few minutes before lighting to clear the line.

A Bunsen burner is commonly used as a heating apparatus. It contains a gas that can be controlled using a gas regulator.

It is necessary to do the following before lighting the Bunsen burner;

Make sure there are no holes in the hose connected to the burner.
Clear the burner area of flammable materials such as solvents and papers.
Check that the gas regulator and air inlet on the burner can be adjusted.

So, it is not necessary to Leave the gas stop open for a few minutes before lighting to clear the line before lighting a Bunsen burner.

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Answer:

the answer is opetuon B

How many moles of MgS2O3 are in 223 g of the compound

Answers

Answer: 1.63 moles

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023* 10^(23)$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}=$

Given mass = 223 g

Molar mass = 136.4

$\text{Number of moles}=(223g)/(136.4g/mol)=1.63moles$

Thus there are 1.63 moles in 223 g of the compound.

Moles of MgS2O3 = 223/molar mass of MgS2O3

= 223/136.42
= 1.634 moles.

Hope this helps!

Determine the number of moles of oxygen atoms in each of the following.1) 4.93 mol H2O2
2) 2.01 mol N2O

Answers

Answer :

Part 1: 4.93 moles of $H_2O_2$ contains 9.86 moles of oxygen atoms.

Part 2: 2.01 moles of $N_2O$ contains 2.01 moles of oxygen atoms.

Explanation :

Part 1: 4.93 mol $H_2O_2$

In 1 mole of $H_2O_2$ , there are 2 atoms of hydrogen and 2 atoms of oxygen.

As, 1 mole of $H_2O_2$ contains 2 moles of oxygen atoms.

So, 4.93 moles of $H_2O_2$ contains $4.93* 2=9.86$ moles of oxygen atoms.

Thus, 4.93 moles of $H_2O_2$ contains 9.86 moles of oxygen atoms.

Part 2: 2.01 mol $N_2O$

In 1 mole of $N_2O$ , there are 2 atoms of nitrogen and 1 atom of oxygen.

As, 1 mole of $N_2O$ contains 1 mole of oxygen atoms.

So, 2.01 moles of $N_2O$ contains $2.01* 1=2.01$ moles of oxygen atoms.

Thus, 2.01 moles of $N_2O$ contains 2.01 moles of oxygen atoms.

Using the balanced equation below,how many grams of carbon dioxide
would be produced from the
complete reaction of 83.7 g carbon
monoxide?
Fe2O3 + 3CO → 2Fe + 3CO2

Answers

131.6 grams of carbon dioxide would be produced from the complete reaction of 83.7 g carbon monoxide.

The balanced chemical equation is given below.

Fe2O3 + 3CO → 2Fe + 3CO2

Calculation,

Since, 28g of carbon dioxide produces 44g of carbon monoxide.

So, 83.7 g of carbon dioxide produces 44×83.7/28 grams

83.7 g of carbon dioxide produces 131.6 grams

What is chemical equation?

The symbolic representation of chemical reaction in which reactant represents in left side and product represents in right side is called chemical equation.

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Answer:131.6 g

Explanation:

Consider the following reaction at equilibrium: NO2(g) + CO(g) = NO(g) + CO2(g) Suppose the volume of the system is decreased at constant temperature, what change will this cause in the system? A shift to produce more NO A shift to produce more CO A shift to produce more NO2 No shift will occur

Answers

Answer: Option (d) is the correct answer.

Explanation:

According to Le Chaltelier's principle, when there occurs any change in an equilibrium reaction then the equilibrium will shift in a direction that will oppose the change.

This means that when pressure is applied on reactant side with more number of moles then the equilibrium will shift on product side that has less number of moles.

For example, $NO_(2)(g) + CO(g) \rightleftharpoons NO(g) + CO_(2)(g)$

Since here, there are same number of moles on both reactant and product side. So, when volume is decreased at a constant temperature in this system then there will occur no change in the equilibrium state.

Thus, we can conclude that in the given when volume of the system is decreased at constant temperature, then no shift will occur.

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