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3 years ago

How does the temperature of the air change with altitude?

1 answer:

8 0

As you increase in elevation, there is less air above you thus the pressure decreases. As the pressure decreases, air molecules spread out further (i.e. air expands) and the temperature decreases. If the humidity is at 100 percent (because it's snowing), the temperature decreases more slowly with height.

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Tin hydrogenooxolate formula

vladimir2022 [97]

Answer:

Tin(IV) Hydrogen Oxalate. Alias: Stannic Hydrogen Oxalate. Formula: Sn(HC2O4)4. Molar Mass: 474.8178. :: Chemistry Applications:: Chemical Elements, Periodic Table.

Explanation:

3 0

3 years ago

Read 2 more answers

Write molecular, complete ionic, and net ionic equations for the reactions that occur, if any, when solutions of the following s

Gnom [1K]

<span>The molecular formula that describes the problem is
2CH3COOH (aq) + Ca(OH)2 (s) ---> Ca(CH3COO)2 (aq) + 2H2O (l)

The net equation is written as follows:
2CH3COOH- (aq) + 2H+ (aq) + Ca(OH)2 (s) ---> Ca2+ (aq) + 2 CH3COO- (aq) + 2H2O (l)
canceling out spectator ions
2H+ (aq) + Ca(OH)2 (s) ---> Ca2+ (aq) + 2 H2O (l)</span>

4 0

3 years ago

A form of electricity which can attract things

qwelly [4]

A form of electricity which can attract things is static electricity. Static electricity is the result of an imbalance between negative and positive charges in an object. These charges built up o the surface of an object until they find a way to e released or discharged causing the attraction of things.

8 0

3 years ago

How many liters of nitrogen dioxide are in 16 g of NO2 at STP?

mojhsa [17]

Answer: 23 liter

Explanation: Im pretty sure

6 0

3 years ago

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