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daser333 [38]
3 years ago
11

Identify one step in the water cycle that is driven by the force of gravity

Chemistry
1 answer:
iren2701 [21]3 years ago
6 0

Answer:

i believe its precipitation??

Explanation:

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What is the formula of Potassium oxide??​
Mashutka [201]
<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ The formula for Potassium oxide is K2O

This is because potassium has a 1+ charge and the oxygen has a 2- charge. Two potassium atoms are required to balance the charge.

<h3><u>✽</u></h3>

➶ Hope This Helps You!

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5 0
3 years ago
Read 2 more answers
What is the difference between autotrophs and heterotroph
IRISSAK [1]
Are there any choices for anwsers

4 0
3 years ago
A toothpick pencil and log are all the same type of wood explain which would be most dense
sineoko [7]
A toothpick will be more dense because of its small mass
6 0
3 years ago
Dissolving potassium chlorate (KClO3) is even more endothermic than potassium chloride.
zhenek [66]

Answer:

The mass of KClO₃ that will absorb the same heat as 5 g of KCl is 3.424 g

Explanation:

Here we have

Heat of solution of KClO₃ = + 41.38 kJ/mol.

Heat of solution of KCl (+17.24 kJ/mol)

Therefore, 1 mole of KCl absorbs +17.24 kJ during dissolution

Molar mass of KCl = 74.5513 g/mol

Molar mass of KClO₃ = 122.55 g/mol

74.5513 g of KCl absorbs +17.24 kJ during dissolution, therefore, 5 g will absorb

\frac{17.24}{74.5513 } \times 5 \, \, kJ \, or  \, 1.156  \, kJ

Therefore the amount of KClO₃ to be dissolved to absorb 1.156 kJ of energy is given by

122.55 g of KClO₃ absorbs + 41.38 kJ, therefore,

\frac{1.156}{41.38} \times 122.55 \,  g = 3.424 \, g

Therefore the mass of KClO₃ that will absorb the same heat as 5 g of KCl = 3.424 g.

5 0
3 years ago
The radioisotope xenon-133 has a half-life of 5.2 days. How much of an 80 gram sample would be left after 20.8 days? PLEASE SHOW
yarga [219]

Answer:

6.73g

Explanation:

T½ = 5.2days

No = 80g

N = ?

T = 20.8days

We'll have to find the disintegration constant first so that we can plug it into the equation that will help us find the mass of the sample after 20.8 days

T½ = In2 / λ

T½ = half life

λ = disintegration constant

λ = In2 / T½

λ = 0.693 / 5.8

λ = 0.119

In(N / No) = -λt

N = final mass of the radioactive sample

No = initial mass of the sample

λ = disintegration constant

t = time for the radioactive decay

In(N/No) = -λt

N / No = e^-λt

N = No(e^-λt)

N = 80 × e^-(0.119 × 20.8)

N = 80 × e^-2.4752

N = 80 × 0.0841

N = 6.728g

The mass of the sample after 20.8 days is approximately 6.73g

6 0
3 years ago
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