Which alkaline earth metal has the smallest molar mass?

Small quantities of h2 gas can be collected by adding hcl to zn. a sample of 195 ml of h2 gas was collected over water at 25 c a

Umnica [9.8K]

<span>15.4 milligrams The ideal gas law is PV = nRT where P = pressure of the gas V = volume of the gas n = number of moles of gas R = Ideal gas constant (8.3144598 L*kPa/(K*mol) ) T = absolute temperature. So let's determine how many moles of gas has been collected. Converting temperature from C to K 273.15 + 25 = 298.15 K Converting pressure from mmHg to kPa 753 mmHg * 0.133322387415 kPa/mmHg = 100.3917577 kPa Taking idea gas equation and solving for n PV = nRT PV/RT = n n = PV/RT Substituting known values n = PV/RT n = (100.3917577 kPa 0.195 L) / (8.3144598 L*kPa/(K*mol) 298.15 K) n = (19.57639275 L*kPa) / (2478.956189 L*kPa/(mol) ) n = 0.007897031 mol So we have a total of 0.007897031 moles of gas particles. Now let's get rid of that percentage that's water vapor. The percentage of water vapor is the vapor pressure of water divided by the total pressure. So 24/753 = 0.03187251 The portion of hydrogen is 1 minus the portion of water vapor. So 1 - 0.03187251 = 0.96812749 So the number of moles of hydrogen is 0.96812749 * 0.007897031 mol = 0.007645332 mol Now just multiple the number of moles by the molar mass of hydrogen gas. Start with the atomic weight. Atomic weight hydrogen = 1.00794 Molar mass H2 = 1.00794 * 2 = 2.01588 g/mol Mass H2 = 2.01588 g/mol * 0.007645332 mol = 0.015412073 g Rounding to 3 significant figures gives 0.0154 g = 15.4 mg</span>

7 0

2 years ago

Read 2 more answers

Convert 2.1 mole of Al2(SO4)3 ionic units to a number of particles.

Semenov [28]

1.26 x 10²⁴particles

Explanation:

Given parameters:

Number of moles of the compound = 2.1 moles

Unknown:

Number of particles contained in the compound = ?

Solution:

A mole can be defined as the amount of a substance that contains Avogadro's number of particles 6.02 x 10²³.

The particles mentioned here can be atoms, molecules, formula units, ions, electrons, protons, neutrons and so forth.

To solve for the number of moles:

Number of particles = number of moles x 6.02 x 10²³

Number of particles = 2.1 x 6.02 x 10²³ = 12.64 x 10²³ = 1.26 x 10²⁴particles

Learn more:

mole calculation brainly.com/question/1841136

#learnwithBrainly

8 0

2 years ago

Some instant cold packs contain ammonium nitrate and a separate pouch of water. When the pack is activated by squeezing to break

lubasha [3.4K]

Answer:

$T_f=-7.81^0C$

Explanation:

Hello,

a) In this case, since the heat associated with the dissolution of ammonium nitrate is positive, such reaction is endothermic as it absorbs heat.

b) Now, for computing the temperature once the dissolution is done, we apply (considering that it is a cooling process):

$T_f=T_0-\frac{\Delta H}{mCp}$

Nonetheless, we should first compute the moles of the mixture as:

$n_{mix}=135.0gH_2O*\frac{1molH_2O}{18gH_2O}+50.0gNH_4NO_3*\frac{1molNH_4NO_3}{80gNH_4NO_3}=8.125mol$

Thus, the total absorbed heat is:

$\Delta H=25.4kJ/mol*8.125mol=206.375kJ$

Now, the temperature is:

$T_f=25.0^0C-\frac{25.4kJ}{(135.0+50.0)g*4.184x10^{-3}kJ/g^0C} \\\\T_f=-7.81^0C$

Best regards.

3 0

2 years ago

A student dissolved 4.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution

mihalych1998 [28]

Answer:

0.08097 grams of nitrate ions are there in the final solution.

Explanation:

Moles of cobalt(II) nitrate ,n= $\frac{4.00 g}{245 g/mol}=0.01633 mol$

Volume of the cobalt(II) nitrate solution, V = 100.0 mL = 0.1 L

$Molarity=\frac{n}{V(L)}$

Let the molarity of the solution be $M_1$

$M_1=\frac{0.01633 mol}{0.1 L}=0.1633 M$

A students then takes 4 .00 mL of $M_1$ solution and dilute it to 275 ml.

$M_1=0.1633 M$

$V_1=4.00 mL$

$M_2=?$ (molarity after dilution)

$V_2=275 mL$ (after dilution)

$M_1V1=M-2V_2$

$M_2=\frac{M_1V_1}{V_2}=\frac{0.1633 M\times 4.00 mL}{275 mL}=0.002375 M$

Molarity of the of solution after dilution is 0.002375 M.

$Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)$

1 mol of cobalt(II) nitrate gives 2 moles of nitrate ions. Then 0.002375 M solution of cobalt (II) nitrate will give:

$[NO_3^{-}]=\frac{2}{1}\times 0.002375 M=0.004750 M$

Moles of nitrate ions = n

Volume of the solution = 275 mL = 0.275 L

Molarity of the nitrate ions = $[NO_3^{-}]=0.004750 M$

$[NO_3^{-}]=\frac{n}{0.275 L}$

n = 0.001306 mol

Mass of 0.001306 moles of nitrate ions:

0.001306 mol × 62 g/mol= 0.08097 g

0.08097 grams of nitrate ions are there in the final solution.

4 0

2 years ago

a block of iron has a mass of 826g. what is the mass of a block of magnesium that has the same volume as the block of iron? the

fiasKO [112]

Answer is: the mass of a block of magnesium is 177.75 grams.

m(Fe) = 826 g.

d(Fe) = 7.9 g/cm³.

1) Calculate volume of iron and magnesium:

d(Fe) = m(Fe) ÷ V(Fe).

V(Fe) = m(Fe) ÷ d(Fe).

V(Fe) = 826 g ÷ 7.9 g/cm³.

V(Fe) = V(Mg) = 104.56 cm³.

2) Calculate mass of magnesium:

m(Mg) = V(Mg) · d(Mg).

m(Mg) = 104.56 g/cm³ · 1.7 g/cm³.

m(Mg) = 177.75 g.

8 0

2 years ago