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RUDIKE [14]
2 years ago
9

A sample of gas occupies 300 mL at 100K. What is its volume when the

Chemistry
1 answer:
iris [78.8K]2 years ago
3 0

If the temperature of the sample of gas increases to the given value, the volume also increases to 600mL.

<h3>What is Charles's law?</h3>

Charles's law states that "the volume occupied by a definite quantity of gas is directly proportional to its absolute temperature.

It is expressed as;

V₁/T₁ = V₂/T₂

Given the data in the question;

  • Initial temperature of gas T₁ = 100K
  • Initial volume of gas V₁ = 300mL
  • Final temperature T₂ = 200K
  • Final volume V₂ = ?

V₁/T₁ = V₂/T₂

V₂ = V₁T₂ / T₁

V₂ = ( 300mL × 200K ) / 100K

V₂ = 60000mLK / 100K

V₂ = 600mL

Therefore, if the temperature of the sample of gas increases to the given value, the volume also increases to 600mL.

Learn more about Charles's law here: brainly.com/question/12835309

#SPJ1

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Answer:


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1) Microwaves are a form of electromagnetic radiation. The same as any wave, they carry energy.


2) The wave length of microwaves are in the range of 0.001 mm to 1 m (shorter than radio waves and longer than infrared)


3) The microwaves of an oven, used to heat food, have a wave length aroun 12 cm.


4) The microwaves transmit energy to the water molecules in the food, by increasing the kinetic energy of water molecules. As result, the water molecules get hotter. Microwaves only penetrate about 1 cm inside the food (a potato for example) and from that the heat is transferred by conduction to the inner parts of the food.

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When a piece of sodium metal is kept in water in chemical equation​
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Explanation:

Sodium metal reacts rapidly with water to form a colourless basic solution of sodium hydroxide (NaOH) and hydrogen gas (H2). The reaction continues even when the solution becomes basic. The resulting solution is basic because of the dissolved hydroxide. 

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3 years ago
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
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Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
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