All categories

3 years ago

23.5 L of h2 is stored at a pressure of 58.7 Kpa what volume would the gas take up at stp

Chemistry

1 answer:

stepan [7]3 years ago

3 0

Answer:- 13.6 L

Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.

Standard pressure is 1 atm that is 101.325 Kpa.

Boyle's law equation is:

$P_1V_1=P_2V_2$

From given information:-

$P_1$ = 58.7 Kpa

$V_1$ = 23.5 L

$P_2$ = 101.325 Kpa

$V_2$ = ?

Let's plug in the values and solve it for final volume.

$58.7Kpa*23.5L=101.325Kpa*V_2$

On rearranging the equation for $V_2$

$V_2=\frac{58.7Kpa*23.5L}{101.325Kpa}$

$V_2$ = 13.6 L

So, the volume of hydrogen gas at STP for the given information is 13.6 L.

You might be interested in

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago

T/F

kifflom [539]

True. Nuclear fusion of hydrogen to form helium occurs naturallyin the sun and other stars. It takes place only at extremely high temperatures.

8 0

3 years ago

True or false solar power take power from wind and uses it to make electricity

quester [9]

The answer is false

3 0

3 years ago

Of the well-characterized elements of group 7A, which would exhibit the following properties? It reacts with the alkali metals (

dezoksy [38]

Answer:

a. fluorine

Explanation:

Fluorine is the element of group 17 and period 2. The electronic configuration of the element is $1s^22s^22p^5$ .

Stable oxidation state = -1 of fluorine as it gains one electron to gain noble gas configuration.

With alkali metals, which have oxidation state of +1 form ionic compound of the form, MX where X is F.

Among the halogens, fluorine forms the most stable halide because of the comparable size of the hydrogen and fluorine. Thus, it is the weakest acid when compared with other hydrogen halides.

Fluorine is the most reactive in the halogen series and thus, combines with most of the elements.

Fluorine forms inter-halogen compounds of form XA only. Example - ClF.

Hence, option a is correct.

8 0

3 years ago

How many moles are in 8.32 times 1024 molecules of CO2

Svetradugi [14.3K]

Answer:

13.8 mol CO₂

General Formulas and Concepts:

Chemistry - Atomic Structure

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

8.32 × 10²⁴ molecules CO₂

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

$8.32 \cdot 10^{24} \ molecules \ CO_2(\frac{1 \ mol \ CO_2}{6.022 \cdot 10^{23} \ molecules \ CO_2} )$ = 13.816 mol CO₂

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

13.816 mol CO₂ ≈ 13.8 mol CO₂

3 0

3 years ago