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lilavasa [31]
3 years ago
13

The sodium chloride molecule breaks apart in water. what does a represent

Chemistry
1 answer:
yarga [219]3 years ago
5 0
What represent ???( the molecules get dissolved in water)..
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Which of these orbital filling diagrams are drawn correctly? (select all that apply)
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A

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The option A is draw well about sha

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The "air bags" that are currently installed in automobiles to prevent injuries in the event of a crash are equipped with sodium
stellarik [79]

Answer:

0.0177 L of nitrogen will be produced

Explanation:

The decomposition reaction of sodium azide will be:

2NaN_{3}(s)--->2Na(s)+3N_{2}(g)

As per the balanced equation two moles of sodium azide will give three moles of nitrogen gas

The molecular weight of sodium azide = 65 g/mol

The mass of sodium azide used = 100 g

The moles of sodium azide used = \frac{mass}{molarmass}=\frac{100}{65}=1.54mol

so 1.54 moles of sodium azide will give = \frac{3X1.54}{2}=2.31mol

the volume will be calculated using ideal gas equation

PV=nRT

Where

P = Pressure = 1.00 atm

V = ?

n = moles = 2.31 mol

R = 0.0821 L atm / mol K

T = 25 °C = 298.15 K

Volume = \frac{P}{nRT}=\frac{1}{2.31X0.0821X298.15}=0.0177L

3 0
3 years ago
LOOK AT THE PICTURE!! i’ll mark as brainliest!!
Lera25 [3.4K]

Answer:

c

Explanation:

8 0
3 years ago
Read 2 more answers
SO
Burka [1]

Answer:

\large \boxed{\text{E) 721 K; B) 86.7 g}}

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL;  V₂ = 435 mL

T₁ = 355 K;     T₂ = ?

b) Calculation

\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}

7 0
3 years ago
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