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3 years ago

What satement is true the about atomic mass?

Chemistry

1 answer:

kirza4 [7]3 years ago

6 0

Answer:

Explanation:

A. It can be greater than the atomic number

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A sample of nitrogen is initially at a pressure of 1.7 kPa, a temperature of -10 C and a volume of 7.5 m3. Then the volume is de

zhannawk [14.2K]

Answer:

$\boxed{\text{2.6 kPa}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}$

Data:

p₁ = 1.7 kPa; V₁ = 7.5 m³; T₁ = -10 °C

p₂ = ?; V₂ = 3.8 m³; T₂ = 200 K

Calculations:

(a) Convert temperature to kelvins

T₁ = (-10 + 273.15) K = 263.15 K

(b) Calculate the pressure

$\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}$

7 0

3 years ago

The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

hammer [34]

Answer:

A) 1059 J/mol

B) 17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

$C_p=C_v+R$

We can determine $C_v$ from above if we make $C_v$ the subject of the formula as:

$C_v$ $=C_p-R$

$C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314$

$C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4$

A).

The formula for calculating change in internal energy is given as:

$dU=C_vdT$

If we integrate above data into the equation; it implies that:

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 1059J/mol$

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 17,920 J/mol$

3 0

4 years ago

On a recent car ride, a student traveled the first 20 kilometers in 5 minutes, the next 60 kilometers in 50 minutes, and the las

inn [45]

Answer:

it is 90 km in one hour it is 90km/h

90/60=1,5

6 0

3 years ago

The crater is the bowl-shaped area around a volcano’s central vent. t/f

Lyrx [107]

If I'm correct, the crater is actually a circular-shaped area around the volcano's central vent. My answer is false

6 0

3 years ago

How many moles of oxygen atoms are present in 0.4 moles of oxygen gas

Agata [3.3K]

Answer:

Each molecule of O2 is made up of 2 oxygen atoms. So 1 mole of O2 molecules is made up of 2 moles of oxygen atoms. Therefore 1 mole of oxygen gas contains 2 moles of oxygen atoms. And 0.4 moles of oxygen gas contains 0.8 moles of oxygen atoms.

4 0

2 years ago

Read 2 more answers