A fact of nature that has been confirmed many times by observation is known as a(n).

What is the mass of NaCl produced when 2.0 moles of chlorine gas

sp2606 [1]

Answer:

Mass = 234 g

Explanation:

Given data:

Mass of NaCl produced = ?

Moles of chlorine = 2.0 mol

Solution:

Chemical equation:

2Na + Cl₂ → 2NaCl

now we will compare the moles of NaCl with chlorine gas.

Cl₂ : NaCl

1 : 2

2.0 : 2/1×2.0 = 4.0 mol

Mass of NaCl;

Mass = number of moles × molar mass

Mass = 4.0 mol × 58.5 g/mol

Mass = 234 g

3 0

3 years ago

What symbol is that plz tell

aleksandr82 [10.1K]

It stands to me it looks like it is a caution

8 0

4 years ago

Read 2 more answers

Which type of land structure forms after a glacier melts

NISA [10]

Mountains and hills form

5 0

3 years ago

Alice and Bob are experimenting with two moles of neon, a monatomic gas, that starts out at conditions of standard temperature a

Archy [21]

Answer:

29273.178 joules have been added to the gas for the entire process.

Explanation:

The specific heats of monoatomic gases, measured in joules per mol-Kelvin, are represented by the following expressions:

Isochoric (Constant volume)

$c_{v} = \frac{3}{2}\cdot R_{u}$ (1)

Isobaric (Constant pressure)

$c_{p} = \frac{5}{2}\cdot R_{u}$ (2)

Where $R_{u}$ is the ideal gas constant, measured in pascal-cubic meters per mol-Kelvin.

Under the assumption of ideal gas, we notice the following relationships:

1) Temperature is directly proportional to pressure.

2) Temperature is directly proportional to volume.

Now we proceed to find all required temperatures below:

(i) Alice heats the gas at constant volume until its pressure is doubled:

$\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}}$ (3)

( $\frac{P_{2}}{P_{1}} = 2$ , $T_{1} = 273.15\,K$ )

$T_{2} = \frac{P_{2}}{P_{1}} \times T_{1}$

$T_{2} = 2\times 273.15\,K$

$T_{2} = 546.3\,K$

(ii) Bob further heats the gas at constant pressure until its volume is doubled:

$\frac{T_{3}}{T_{2}} =\frac{V_{3}}{V_{2}}$ (4)

( $\frac{V_{3}}{V_{2}} = 2$ , $T_{2} = 546.3\,K$ )

$T_{3} = \frac{V_{3}}{V_{2}}\times T_{2}$

$T_{3} = 2\times 546.3\,K$

$T_{3} = 1092.6\,K$

Finally, the heat added to the gas ( $Q$ ), measured in joules, for the entire process is:

$Q = n\cdot [c_{v}\cdot (T_{2}-T_{1})+c_{p}\cdot (T_{3}-T_{2})]$ (5)

If we know that $R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}$ , $n = 2\,mol$ , $T_{1} = 273.15\,K$ , $T_{2} = 546.3\,K$ and $T_{3} = 1092.6\,K$ , the heat added to the gas for the entire process is:

$c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{v} = 12.471\,\frac{J}{mol\cdot K}$

$c_{p} = \frac{5}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{p} = 20.785\,\frac{J}{mol\cdot K}$

$Q = (2\,mol)\cdot \left[\left(12.471\,\frac{J}{mol\cdot K} \right)\cdot (536.3\,K-273.15\,K)+\left(20.785\,\frac{J}{mol\cdot K} \right)\cdot (1092.6\,K-546.3\,K )\right]$

$Q = 29273.178\,J$

29273.178 joules have been added to the gas for the entire process.

6 0

3 years ago

A 642 mL sample of oxygen gas at 23.5°C and 795 mm Hg, is heated to 31.7°C and the volume of the gas expands to 957 mL. What is

Ratling [72]

We can use the combined gas law equation to find the new pressure of the gas.
$\frac{P1V1}{T1} = \frac{P2V2}{T2}$
where P - pressure
V - volume
T - temperature
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
P1 - 795 mm Hg x 0.0013 atm/ mm Hg = 1.033 atm
T1 - 23.5 °C + 273 = 296.5 K
T2 - 31.7 °C + 273 = 304.7 K
substituting the values in the equation
$\frac{1.033 atm *642 mL}{296.5K}= \frac{P*957mL}{304.7K}$
P = 0.712 atm
the answer closest to this value is A) 0.723 atm
therefore answer is
A) 0.723 atm

8 0

4 years ago