How is gas different from a liquid

Chemistry

2 answers:

vlada-n [284]3 years ago

6 0

The gas is different from a liquid because the gas can escape a liquid doesn't

satela [25.4K]3 years ago

4 0

The atoms and molecules in gases are much more spread out than in solids orliquids. They vibrate and move freely at high speeds. A gas will fill any container, but if the container is not sealed, the gas will escape. Gas can be compressed much more easily than a liquid or solid

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CaC2 + 2H2O ⟶ C2H2 + Ca(OH)2 How many moles of water are needed to produce 56.8g C2H2?

pshichka [43]

Amount of moles of water is 4.36 mol

Explanation: M(C2H2) = 2·12.01 + 2·1.008 = 26.036 g/mol

Amount of substance is n= m/M= 56.8 g / 26.036 g/mol= 2.18159 mol

Water is needed 2·2.18 mol = 4.36 mol

6 0

3 years ago

Lab: Limiting Reactant and Percent Yield

creativ13 [48]

Answer : If a substance is the limiting reactant, then it limits the formation of products because in the reaction it is present in limited amount.

Explanation :

While observing a chemical reaction, we can tell about whether a reactant is limiting or excess.

Step 1 : first write the chemical reaction and then balanced the chemical equation.

$C_2H_4+H_2\rightarrow C_2H_6$

Step 2 : convert the given masses into the moles if mass of $C_2H_4$ is 10.5 g and molar mass of $C_2H_4$ is 28 g/mole and the mass of hydrogen is 0.40 g and molar mass of hydrogen is 2 g/mole.

$\text{moles of }C_{2}H_{4}=\frac{10.5g}{28g/mole}=0.375moles$

$\text{moles of }H_{2}=\frac{0.40g}{2g/mole}=0.20moles$

Step 3 : Now we have to determine the limiting reagent and excess reagent.

$\text{ moles of }C_{2}H_{4}\text{ in excess}=0.375-0.20=0.175\text{moles}$

Now we conclude that $C_{2}H_{4}$ is the limiting reagent and hydrogen is an excess reagent.

Hypothesis :

Limiting reagent : It is the reagent in the chemical reaction that is totally consumed when the chemical reaction is complete. Limiting reagent limits the formation of products.

7 0

2 years ago

Read 2 more answers

What is the ph of a 0.20 M solution of NH4Cl? [Kb(NH3) = 1.8 x 10^-5]

Pavlova-9 [17]

"NH4+ <----> NH3 + H+

The constant of this equilibrium is: K = Kw / Kb = 1 x 10^-14 / 1.8 x 10^-5 =5.56 x 10^-10

5.56 x 10^-10 = x^2 / 0.20-x

x = [H+] =1.1 x 10^-5 M

pH = 5.0"

8 0

3 years ago