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2 years ago

During the charging process, the allied liquid becomes gas. What happens to particles in the liquid

1 answer:

8 0

Answer: The particles need energy to overcome the attractions between them. As the liquid gets warmer more particles have sufficient energy to escape from the liquid. Eventually even particles in the middle of the liquid form bubbles of gas in the liquid.

Explanation:

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How many grams of H, are needed to react with 2.75 g of N,?

Elena-2011 [213]

Answer:

0.6 grams of hydrogen are needed to react with 2.75 g of nitrogen.

Explanation:

When hydrogen and nitrogen react they form ammonia.

Chemical equation:

N₂ + 3H₂ → 2NH₃

Given mass of nitrogen = 2.75 g

Number of moles of nitrogen:

Number of moles = mass/ molar mass

Number of moles = 2.75 g / 28 g/mol

Number of moles = 0.098 mol

Now we will compare the moles of nitrogen with hydrogen from balance chemical equation:

N₂ : H₂

1 : 3

0.098 : 3×0.098 = 0.3 mol

Mass of hydrogen:

Mass = number of moles × molar mass

Mass = 0.3 mol × 2 g/mol

Mass = 0.6 g

6 0

3 years ago

How dose heat travel between objects that are touching one another?

11Alexandr11 [23.1K]

Through the process of conduction, heat van be trnafered between two solids.

3 0

3 years ago

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Wide shield-like mountains are formed from what type of lava?

Kaylis [27]

Thick liquid lava
It's right

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2 years ago

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A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch

kozerog [31]

Answer:

$\large \boxed{\text{4.63 atm}}$

Explanation:

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

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

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C

p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}$