All categories

3 years ago

Please balance xe+f2=xef6

Chemistry

1 answer:

Marysya12 [62]3 years ago

8 0

If’s chemistry it would be like this- Xe=1 and F=2/Xe=1 and F=6 so, to balance Xe=1 and F=[3x2]/Xe=1 and F=6.

You might be interested in

What do you think would happen to a rock after billion years?

ANEK [815]

Answer:

Explanation:

It probably would have been broken down to smaller pieces and spread out around the world. Hope this help! :)

3 0

3 years ago

What is chemical change

Igoryamba

Answer:

Chemical change is a chemical process that occurs when a substance combines with another to create an entirely new substance, this process is also called chemical synthesis

Explanation:

8 0

3 years ago

When may a scientific theory be revised?

Anika [276]

Answer:

it can be revised at any time.

Explanation:

6 0

4 years ago

Read 2 more answers

The equation 2N2(g) + 3H2(g) = 2NH3(g) + energy shows the

Alexandra [31]

The equilibrium shifts to the side having the lesser volume or number of moles

Explanation:

Reaction 2N₂ $g_{123}$ + 3H₂ $g_{123}$ ⇄ 2NH₃ $g_{123}$

This reaction, haber process is completely a combination of gaseous molecules.

An increase in pressure will shift the equilibrium position to the right hand side i.e the side with the lesser volume or number of moles.

Number of moles

Reactants Products

5 2

The product side has a lesser number of moles and the equilibrium shifts to the right hand side.

Learn more:

Equilibrium brainly.com/question/5877801

#learnwithBrainly

3 0

3 years ago

Iron-59 is a beta emitter with a half-life of 44.5 days. If a sample initially contains 180 mg of iron-59, how much iron-59 is l

Jobisdone [24]

Answer:

2.79 mg iron-59 is left in the sample after 267 days.

Explanation:

Given that:

Half life = 44.5 days

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{44.5}\ days^{-1}$

The rate constant, k = 0.0156 days⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration = 180 mg

Time = 267 days

So,

$[A_t]=180\ mg\times e^{-0.0156\times 267}$

$[A_t]=180\times e^{-0.0156\times 267}\ mg$

$[A_t]=2.79\ mg$

<u>2.79 mg iron-59 is left in the sample after 267 days.</u>

5 0

3 years ago