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3 years ago

400 mL of gas is contained at 300 mmHg and 0 °C. What will its volume be at 140 mmHg and 100 °C? 0°C 100°C

Chemistry

1 answer:

PIT_PIT [208]3 years ago

7 0

Answer:

1171.12 mL

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

V1 = initial volume (milliliters)

V2 = final volume (milliliters)

T1 = initial temperature (Kelvin)

T2 = final temperature (Kelvin)

According to the information provided in this question:

P1 = 300 mmHg

P2 = 140 mmHg

V1 = 400 mL

V2 = ?

T1 = 0°C = 273K

T2 = 100°C = 100 + 273 = 373K

Using P1V1/T1 = P2V2/T2

300 × 400/273 = 140 × V2/373

120000/273 = 140V2/373

120000 × 373 = 273 × 140V2

44760000 = 38220V2

V2 = 44760000 ÷ 38220

V2 = 1171.115

The new volume is 1171.12 mL

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For the following reactions, write a balanced equation using half-reactions and calculate the voltage to be expected.

Iteru [2.4K]

a) $2Na(s) +2H_2O(I)$ → $2Na^+(aq) + OH^-(aq)+ H_2(g)$

b) $2Ag (s) +2H^(aq)$ → $2 Ag^+ (aq) +H_2(g)$

<h3>What are half-reactions?</h3>

The half-reaction method is a way to balance redox reactions. It involves breaking the overall equation down into an oxidation part and a reduction part.

$2Na(s) +2H_2O(I)$ → $2Na^+(aq) + OH^-(aq)+ H_2(g)$

$E^0 cell = E^0 (reduction) - E^0 (oxidation)$

= $E^0(\frac{H_2O}{H_2}, OH^-) -E^0(Na^+/Na)$

= -0.83 - (-2.71) =1.88V

$2Ag (s) +2H^(aq)$ → $2 Ag^+ (aq) +H_2(g)$

$E^0cell$ = $E^0(H^+/H_2) -E^0(Ag^+/Ag)$

$E^0cell$ =-0. - (0.8) =-0.8V

Learn more about the half-reactions here:

https://brainly.in/question/18053421

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8 0

2 years ago

Dumbledore decides to gives a surprise demonstration. He starts with a hydrate of Na2CO3 which has a mass of 4.31 g before heati

vovikov84 [41]

Answer:

Na₂CO₃.2H₂O

Explanation:

For the hydrated compound, let us denote is by Na₂CO₃.xH₂O

The unknown is the value of x which is the amount of water of crystallisation.

Given values:

Starting mass of hydrate i.e Na₂CO₃.xH₂O = 4.31g

Mass after heating (Na₂CO₃) = 3.22g

Mass of the water of crystallisation = (4.31-3.22)g = 1.09g

To determine the integer x, we find the number of moles of the anhydrous Na₂CO₃ and that of the water of crystallisation:

Number of moles = $\frac{mass }{molar mass }$

Molar mass of Na₂CO₃ =[(23x2) + 12 + (16x3)] = 106gmol⁻¹

Molar mass of H₂O = [(1x2) + (16)] = 18gmol⁻¹

Number of moles of Na₂CO₃ = $\frac{3.22}{106}$ = 0.03mole

Number of moles of H₂O = $\frac{1.09}{18}$ = 0.06mole

From the obtained number of moles:

Na₂CO₃ H₂O

0.03 0.06

Simplest

Ratio 0.03/0.03 0.03/0.06

1 2

Therefore, x = 2

7 0

4 years ago

If 1.20 moles of an ideal gas occupy a volume of 18.2 l at a pressure of 1.80 atm, what is the temperature of the gas, in degree

scoundrel [369]

We can calculate for temperature by assuming the equation for ideal gas law:

P V = n R T

Where,

P = pressure = 1.80 atm

V = volume = 18.2 L

n = number of moles = 1.20 moles

R = gas constant = 0.08205746 L atm / mol K

Substituting to the given equation:

T = P V / n R

T = (1.8 atm * 18.2 L) / (1.2 moles * 0.08205746 L atm / mol K)

T = 332.70 K

We can convert K unit to ˚C unit by subtracting 273.15 to Kelvin, therefore

T = 59.55 ˚<span>C</span>

6 0

4 years ago

PLEASE HELP!!!!! I need help with this question!

Ivahew [28]

Answer:

Exothermic Reaction

Explanation:

Its a combustion reaction and they are always exothermic in nature.

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MrMuchimi

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