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Ede4ka [16]
3 years ago
13

Find the volume of a gas at STP, if its volume is 80.0 mL at 109 kPa and -12.5°C.​

Chemistry
2 answers:
ipn [44]3 years ago
8 0

<u>Answer:</u> The volume when the pressure and temperature has changed is 90.21 mL

<u>Explanation:</u>

<u>At STP:</u>

The temperature at this condition is taken as 273 K

The pressure at this condition is taken as 1 atm or 101.3 kPa.

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

We are given:

P_1=101.3kPa\\V_1=?mL\\T_1=273K\\P_2=109kPa\\V_2=80.0mL\\T_2=-12.5^oC=[-12.5+273]K=260.5K

Putting values in above equation, we get:

\frac{101.3kPa\times V_1}{273K}=\frac{109kPa\times 80}{260.5K}\\\\V_1=\frac{109\times 80\times 273}{101.3\times 260.5}=90.21mL

Hence, the volume when the pressure and temperature has changed is 90.21 mL

exis [7]3 years ago
5 0

Answer:

= 913.84 mL

Explanation:

Using the combined gas laws

P1V1/T1 = P2V2/T2

At standard temperature and pressure. the pressure is 10 kPa, while the temperature is 273 K.

V1 = 80.0 mL

P1 = 109 kPa

T1 = -12.5 + 273 = 260.5 K

P2 = 10 kPa

V2 = ?

T2 = 273 K

Therefore;

V2 = P1V1T2/P2T1

     = (109 kPa × 80 mL × 273 K)/(10 kPa× 260.5 K)

     <u>= 913.84 mL</u>

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In a titration of hno3, you add a few drops of phenolphthalein indicator to 50.00 ml of acid in a flask. You quickly add 20.00 m
bekas [8.4K]

Answer:

The HNO3 solution has a concentration of 0.07 M

Explanation:

<u>Step 1:</u> find a balanced equation

HNO3 (aq) + NaOH (aq) → NaNO3 (aq) + H2O (l)

⇒ for 1 mole of HNO3 reacted, there will also react 1 mole of NaOH, and be produced 1 mole of NaNO3 and 1 mole of H2O, since the ratio is 1:1

<u>Step 2:</u> Calculating moles

Since we know that for 1 mole of HNO3 there will react 1 mole of NaOH, we can calculate the number of moleNaOH

⇒ Concentration = mole / volume

⇒ 0.210 = mole / ((20 + 7.23 ml) *10^-3)

mole = 0.005733 mole NaOH  = 0.005733 mole HNO3

<u>Step 3:</u> Calculating the concentration of HNO3

Concentration = mole / volume

C(HNO3) = 0.005733 mole / ((50 + 30 ml) *10^-3)

C(HNO3) = 0.07 M

The HNO3 solution has a concentration of 0.07 M

To control this we can calculate through the following formule:

0.02723L x 0.21 M x ( 1mol HNO3 / 1 mol NaOH) x (1/ 0.08L) = 0.07M

8 0
3 years ago
When sulfur burns in air, it forms sulfur dioxide as shown by the equation below.what volume of SO2 is produced when 2.35 g of s
Naddik [55]

1.64 L of sulfur dioxide (SO₂)

Explanation:

We have the following chemical reaction:

S (s) + O₂ (g) → SO₂ (g)

First we calculate the number of moles of sulfur (S):

number of moles = mass / molar weight

number of moles of sulfur = 2.35 / 32 = 0.0734 moles

Looking at the chemical reaction we see that 1 moles of sulfur (S) produces 1 moles of sulfur dioxide (SO₂), so 0.0734 moles of sulfur will produce 0.0734 moles of sulfur dioxide (SO₂).

To calculate the volume of sulfur dioxide (SO₂), assuming that the sulfur dioxide is behaving as an ideal gas and the we determine the gas volume under standard temperature and pressure conditions, we use the following formula:

number of moles = volume / 22.4 (L/mole)

volume = number of moles × 22.4

volume of SO₂ = 0.0734 × 22.4 = 1.64 L

Learn more about:

molar volume

brainly.com/question/11160940

brainly.com/question/506048

#learnwithBrainly

PS: I appreciate that you took the time and effort to write the chemical equation in a readable way. This makes the question to be very rare :D

5 0
3 years ago
A mixture of ethyl acetate vapour and air has a relative saturation of 50% at 303 K and a total pressure of 100 kPa. If the vapo
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Answer : The correct option is, (b) 0.087

Explanation :

The formula used for relative saturation is:

\text{Relative saturation}=\frac{P_A}{P_A^o}

where,

P_A = partial pressure of ethyl acetate

P_A^o = vapor pressure of ethyl acetate

Given:

Relative saturation = 50 % = 0.5

Vapor pressure of ethyl acetate = 16 kPa

Now put all the given values in the above formula, we get:

0.5=\frac{P_A}{16kPa}

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Now we have to calculate the molar saturation.

The formula used for molar saturation is:

\text{Molar saturation}=\frac{P_{vapor}}{P_{\text{vapor free}}}

and,

P(vapor free) = Total pressure - Vapor pressure

P(vapor) = P_A = 8 kPa

So,

P(vapor free) = 100 kPa - 8 kPa = 92 kPa

The molar saturation will be:

\text{Molar saturation}=\frac{P_{vapor}}{P_{\text{vapor free}}}

\text{Molar saturation}=\frac{8kPa}{92kPa}=0.087

Therefore, the molar saturation is 0.087

5 0
3 years ago
Pls help and look at both ily
bekas [8.4K]

Answer:

No

Explanation:

(I'm assuming this is a decomposition reaction)

The model is incomplete as a part of the reactants (sulfur) is not present in the products

6 0
3 years ago
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