Ask question

Ask question

All categories

gayaneshka [121]

2 years ago

7

If 10.0 liters of h2(g) at STP is heated to a temperature of 546 K, pressure remaining constant, the new volume of the gas will

be

1 answer:

Effectus [21]2 years ago

3 0

Answer:

20L

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 10L

Initial Temperature (T1) = stp = 273K

Final temperature (T2) = 546K

Final volume (V2) =..?

The new volume of the gas can be obtained by using the Charles' law equation as shown below:

V1/T1 = V2/T2

10/273 = V2/546

Cross multiply to express in linear form

273 x V2 = 10 x 546

Divide both side by 273

V2 = (10 x 546) / 273

V2 = 20L

Therefore, the new volume of the gas is 20L.

You might be interested in

According to the equation above, how many moles of potassium chlorate, KClO3, must be decomposed to generate 1.0 L of O2 gas at

nignag [31]

Answer:

Moles of potassium chlorate = 0.02976 moles

Explanation:

At standard pressure and temperature,

22.4 L of a gas consists of 1 mole

Thus, given, volume of $O_2$ = 1.0 L

So,

1 L of a gas consists of $\frac{1}{22.4}$ mole

Moles of oxygen gas = 0.04464 moles

The reaction is shown below as:-

$2KClO_3\rightarrow 2KCl+3O_2$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

1 mole of oxygen gas are produced when $\frac{2}{3}$ moles of potassium chlorate undergoes reaction.

Thus,

0.04464 mole of oxygen gas are produced when $\frac{2}{3}\times 0.04464$ moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate = 0.02976 moles</u>

6 0

2 years ago

At the equivalence point of a KHP/NaOH titration, you have added enough OH- to react with all of the HP- such that the only spec

Nataly_w [17]

Explanation:

The given data is as follows.

$[P^{2-}]$ = 0.042 M, $K_{a}$ for $HP^{-} = 3.9 \times 10^{-6}$

According to the given situation $P^{2-}$ acts as a base.The reaction equation will be as follows.

$P^{2-} + H_{2}O \rightleftharpoons HP^{-} + OH^{-}$

Relation between $K_{b}$ and $K_{a}$ are as follows.

$K_{a} \times K_{b} = K_{w}$

$K_{b} = \frac{1 \times 10^{-14}}{K_{a}}$

= $\frac{1 \times 10^{-14}}{3.9 \times 10^{-6}}}$

= $2.6 \times 10^{-9}$

Also, $K_{b} = \frac{[HP^{-}][OH^{-}]}{P^{2-}}$

Let us take $[OH^{-}] = [HP^{-}]$ = x

So, $K_{b} = \frac{[HP^{-}][OH^{-}]}{P^{2-}}$

$2.6 \times 10^{-9} = \frac{x \times x}{0.042}$

x = $1.04 \times 10^{-5}$

$[OH^{-}] = [HP^{-}]$ = $1.04 \times 10^{-5}$

pOH = - log $[OH^{-}]$

= - log ( $1.04 \times 10^{-5}$ )

= 4.99

As it is known that pH + pOH = 14

so, pH + 4.99 = 14

pH = 9.01

Thus, we can conclude that pH of the solution is 9.01.

4 0

2 years ago

A titration is performed to determine the amount of sulfuric acid, H2SO4, in a 6.5 mL sample taken from car battery. About 50 mL

Orlov [11]

Answer: The molar concentration of sulfuric acid in the original sample is 1.943 M

Explanation:

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?\\V_1=56.5mL\\n_2=1\\M_2=0.5824M\\V_2=43.37mL$

Putting values in above equation, we get:

$2\times M_1\times 56.5=1\times 0.5824\times 43.37$

$M_1=0.2235$

Now to calculate the molarity of original solution:

$M_1\times 6.5=0.2235\times 56.5$

$M_1=1.943$

Thus the molar concentration of sulfuric acid in the original sample is 1.943 M

5 0

2 years ago

A reaction mixture initially contains 0.86 atm NO and 0.86 atm SO3. Determine the equilibrium pressure of NO2 if Kp for the reac

dlinn [17]

Explanation:

Reaction equation for this reaction is as follows.

$NO(g) + SO_{3}(g) \rightarrow NO_{2}(g) + SO_{2}(g)$

It is given that $K_{p}$ = 0.0118.

According to the ICE table,

$NO(g) + SO_{3}(g) \rightarrow NO_{2}(g) + SO_{2}(g)$

Initial: 0.86 0.86 0 0

Change: -x -x +x +x

Equilibrium: 0.86 - x 0.86 - x x x

Hence, value of $K_{p}$ will be calculated as follows.

$K_{p} = \frac{P_{NO_{2}} \times P_{SO_{3}}}{P_{NO} \times P_{SO_{3}}}$

0.0118 = $\frac{x \times x}{(0.86 - x)^{2}}$

x = 0.084 atm

Thus, we can conclude that $P_{NO_{2}}$ is 0.084 atm.

4 0

2 years ago

Aluminum–lithium (Al–Li) alloys have been developed by the aircraft industry to reduce the weight and improve the performance of

gtnhenbr [62]

Answer:

The concentration of Li (in wt%) is 3,47g/mol

Explanation:

To obtain the 2,42g/cm³ of density:

2,42g/cm³ = 2,71g/cm³X + 0,534g/cm³Y <em>(1)</em>

<em>Where X is molar fraction of Al and Y is molar fraction of Li.</em>

X + Y = 1 <em>(2)</em>

Replacing (2) in (1):

Y = 0,13

Thus, X = 0,87

The weight of Al and Li is:

0,87*26,98g/mol = 23,4726 g of aluminium

0,13*6,941g/mol = 0,84383 g of lithium

The concentration of Li (in wt%) is:

0,84383g/(0,84383g+23,4726g) ×100= <em>3,47%</em>

6 0

2 years ago