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3 years ago

What volume does 2800mg of nitrogen gas occupy at 98Kpa and -10 C?

Chemistry

1 answer:

Evgen [1.6K]3 years ago

6 0

Answer:

2.23L

Explanation:

Using the general gas law as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas constant (0.0821 Latm/molK)

T = temperature (K)

According to the information provided in this question,

Mass of nitrogen = 2800mg

Since 1g = 1000mg

2800mg = 2800/1000

= 2.8g

Using mole = mass ÷ molar mass

Molar mass of nitrogen gas (N2) = 28g/mol

mole = 2.8/28

mole of N2 gas = 0.1mol

Pressure = 98kPa

1 kilopascal (kPa) = 1000pascal (pa)

98kPa = 98 × 1000

= 98000pascal

Since 1 Pascal = 9.869 × 10^-6 atmosphere (atm)

98000pascal = 98000 × 9.869 × 10^-6

= 0.967atm

Temperature = -10°C

Kelvin = °C + 273

Kelvin = -10 + 273

K = 263K

Hence, using PV = nRT

0.967 × V = 0.1 × 0.0821 × 263

0.967V = 2.159

V = 2.159/0.967

V = 2.23

Volume of Nitrogen gas = 2.23 L

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Equal masses of substance A and substance B are at 25°C. Substance A has a lower specific heat capacity than substance B. If 50

liraira [26]

Dang bro that stuff is really hard I’m defiantly on a lower grade lol

6 0

3 years ago

4.45 kcal of heat was added to increase the temperature of a sample of water from 23.0 °C to 57.8 °C. Calculate

Alona [7]

Answer:

m = 4450 g

Explanation:

Given data:

Amount of heat added = 4.45 Kcal ( 4.45 kcal ×1000 cal/ 1kcal = 4450 cal)

Initial temperature = 23.0°C

Final temperature = 57.8°C

Specific heat capacity of water = 1 cal/g.°C

Mass of water in gram = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 57.8°C - 23.0°C

ΔT = 34.8°C

4450 cal = m × 1 cal/g.°C × 34.8°C

m = 4450 cal / 1 cal/g

m = 4450 g

4 0

3 years ago

How many pounds in a 3.00 l bottle of drinking water?

slava [35]

Answer:

6.61 Pounds

Solution:

Step 1: Calculate Mass of Water as;

Density = Mass ÷ Volume

Solving for Mass,

Mass = Density × Volume ------ (1)

As,

Density of Water = 1 g.cm⁻³

And,

3 L of Water = 3000 cm³

Putting values in equation 1,

Mass = 1 g.cm⁻³ × 3000 cm³

Mass = 3000 g

Step 2: Convert Grams into Pounds;

As,

1 Gram = 0.002204 Pounds

So,

3000 Grams = X Pounds

Solving for X,

X = (3000 Grams × 0.002204 Pounds) ÷ 1 Gram

X = 6.61 Pounds

6 0

4 years ago

N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol

docker41 [41]

Answer:

Explanation:

Recall that ΔH is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:

$\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)$

Therefore, from the chemical equation, we have that:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} + \Delta H^\circ_f \text{ H$_2$O} \right] -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}$

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}$

In conclusion, our answer is A.

5 0

3 years ago

What is the magnitude of earthquake took place in manila? A.5.2 B.5.3 C.5.4 D.5.5

stepan [7]

Answer:

i believe its 5.2

Explanation:

6 0

3 years ago