The number of grams of H2 in 1470 mL of H2 gas.

Chemistry

1 answer:

Lorico [155]2 years ago

5 0

Answer:

0.1313 g.

Explanation:

It is known that at STP, 1.0 mole of ideal gas occupies 22.4 L.

Suppose that hydrogen behaves ideally and at STP conditions.

Using cross multiplication:

1.0 mol of hydrogen occupies → 22.4 L.

??? mol of hydrogen occupies → 1.47 L.

∴ The no. of moles of hydrogen that occupies 1.47 L = (1.0 mol)(1.47 L)/(22.4 L) = 6.563 x 10⁻² mol.

Now, we can get the no. of grams of hydrogen in 6.563 x 10⁻² mol:

The no. of grams of hydrogen = no. of hydrogen moles x molar mass of hydrogen = (6.563 x 10⁻² mol)(2.0 g/mol) = 0.1313 g.

You might be interested in

Which of the following statements explains why atoms are always neutral in charge

alex41 [277]

Answer see because there is a lack of potassium peanuts and walnuts also help the gravitational pull moons rocks different one but because of this energy conversion is the after effect of the camera being shot

7 0

2 years ago

The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.

azamat

Answer:

(a) $m_{gold}=7.322g$

(b)

$V_{gold}=0.379cm^3$

$V_{copper}=0.122cm^3$

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign as shown below:

$m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g$

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

$V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3$