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2 years ago

A sample of helium gas occupies 12.9 L at 315 K and 1.20 atm.

Chemistry

1 answer:

julsineya [31]2 years ago

5 0

Answer:

There are 0,6 moles of helium.

Explanation:

We use the ideal gas formula, we use the gas constant R = 0.082 l atm / K mol and we solve for n (number of moles) of the formula:

PV=nRT --> n= PV/RT

n= 1,20 atm x 12,9 L/ 0.082 l atm / K mol x 315 K= <em>0,599303135 moles</em>

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Catherine has some seawater. She wants to separate pure water from the seawater.

Flura [38]

Answer:

the change is evaporation

Explanation:

the water heats up at the surface of the water and evaporates

4 0

2 years ago

Read 2 more answers

Suppose a container holds 1000 hydrogen molecules (H2) and 1000 oxygen molecules (O2) that react to form water. How many water m

mezya [45]

500 water molecules and the remaining 500 O2 molecules. Remember the ratio of H to O in H2O.

8 0

2 years ago

If 3.18 x 10^23 atoms of iron react with 67.2 L of chlorine gas at STP, what is the maximum

spin [16.1K]

84.34 grams of grams of iron (III) chloride that can be produced is maximum because Fe is the limiting reagent in this reaction and chlorine gas is excess reagent.

Explanation:

Balanced chemical equation:

2 Fe + 3 Cl2 → 2 FeCl3

DATA GIVEN:

iron = atoms

mass of chlorine gas = 67.2 liters

mass of FeCl3 = ?

number of moles of iron will be calculated as

number of moles = $\frac{total number of atoms}{Avagaro's number}$

number of moles = $\frac{3.18 x 10^23}{6.022x 10^23}$

number of moles = 0.52 moles of iron

moles of chlorine gas

number of moles = $\frac{mass}{molar mass of 1 mole}$

Putting the values in the equation:

n = $\frac{67200}{70.96}$ (atomic mass of chlorine gas = 70.96 grams/mole)

= 947.01 moles

Fe is the limiting reagent so

2 moles of Fe gives 2 moles of FeCl3

0.52 moles of Fe will give

$\frac{2}{2}$ = $\frac{x}{0.52}$

0.52 moles of FeCl3 is formed.

to convert it into grams:

mass = n X atomic mass

= 0.52 x 162.2 (atomic mass of FeCl3 is 162.2grams/mole)

<h3> = 84.34 grams </h3>

3 0

2 years ago

Why does 4.03/0.0000035 = 1.2 x 106, instead of a different number of significant figures?

jasenka [17]

Explanations:- As per the significant figures rule, In multiplication and division, we go with least number of sig figs.

4.03 has three sig figs where as 0.0000035 has two sig figs only, The zeros in this number are not sig figs as they are just holding the place values. As the least number of sig figs here is two, the answer needs to be reported with two sig figs only.

$\frac{4.03}{0.0000035}=1.2*10^6$

4 0

2 years ago

A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is

Colt1911 [192]

Answer:

$M_{Na^+}=1.36M$

$M_{Br^-}=1.58M$

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

$n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+$

Once we've got the moles we compute the final volume via:

$V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L$

Thus, the molarity of the sodium atoms turn out into:

$M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M$

Now, we perform the same procedure but now for the bromide ions:

$n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-$

Finally, its molarity results:

$M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M$

Best regards.

7 0

2 years ago