All categories

3 years ago

What gas has formed when bubels form when water is boiled?

1 answer:

4 0

The gas has formed when bubbles form when water is boiled are hydrogen and oxygen gas. They came from the reaction 2H2O -> 2H2 + O2 were O2 and N2 will separate after heat treatment.

You might be interested in

What causes atoms to form covalent bonds?

navik [9.2K]

Answer:

to attain stability atoms form covalent bonds

7 0

1 year ago

A home remodeling company charges $10 per square yard for installing new carpets. What is the cost for installing a carpet for a

balu736 [363]

Answer:

$200.0 is the cost for installing a carpet for a room that measures 12.0 feet by 15.0 feet.

Explanation:

Length of the room = l = 15.0 ft = $\frac{15.0}{3} yard=5.0 yard$

Width of the room = w = 12.0 ft = $\frac{12.0}{3} yard=4.0 yard$

(1 yard = 3 feet)

Area of the room = l × w = $4.0 yard\times 5.0 yard = 20.0 yard^2$

Cost for installing a carpet for a room = $10 per square yard

Cost for installing a carpet for a room wit area 20.0 square yard :

$20.0 yard^2\times $10/ yard^2=$200.0$

$200.0 is the cost for installing a carpet for a room that measures 12.0 feet by 15.0 feet.

7 0

3 years ago

Use the table to determine the number of protons for each atom. Then, choose the statement below that is true about the two atom

Lilit [14]

C
The elements are Lithium and Beryllium— just look at the mass number on the periodic table

5 0

2 years ago

If it requires 75.0mL of 0.500M NaOH to neutralize 165.0 mL of an hcl solution what is the concentration of the hcl solution

Kaylis [27]

Answer:

0.027 M HCl

Explanation:

The chemical equation of the neutralization is:

1 NaOH + 1 HCl -> 1 H2O + 1 NaCl

Because the ratio of NaOH and HCl is 1:1 you can use the M1V1=M2V2 formula.

(75 mL)(0.5 M NaOH) = (165 mL)(M HCl)

It requires 0.027 M HCl.

7 0

3 years ago

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

Answer: The value of $K_{eq}$ is $4.84\times 10^{-5}$

Explanation:

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0280 0.0120

At eqllm: 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

4 0

3 years ago