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vova2212 [387]
3 years ago
15

A 50 kg object _____.

Chemistry
1 answer:
stich3 [128]3 years ago
8 0

Answer:

will have more mass the moon than the earth

Explanation:

hope this helps

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alexandr1967 [171]
The percentage of hydrogen in C7H18 is calculated as follows:
[18/(12*7+1*8)]*100=18%
The amount of hydrogen in 5.2moles is given by:(18/100)*5.2=0.94moles
6 0
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Which chemical equation is unbalanced? c o2 right arrow. co2 sr o2 right arrow. 2sro 6h2 3o2 right arrow. 6h2o h2 h2 o2 right ar
netineya [11]

The unbalanced equation is one in which the moles of atoms are not equal on both sides of the reaction.

<h3>What is a balanced chemical equation?</h3>

A balanced chemical equation is one in wgich the moles of atoms in the reactants side is equal to the moles of atoms on the product side.

The given equations of reaction is not clearly stated.

Therefore, the unbalanced equation is one in which the moles of atoms are not equal on both sides of the reaction.

Learn more about balanced equations at: brainly.com/question/11904811

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2 years ago
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3 years ago
A solution is made by dissolving 0.565 g of potassium nitrate in enough water to make up 250. mL of solution. What is the molari
aalyn [17]
<h3>Molar mass of Potassium Nitrate:-</h3>

\\ \large\sf\longmapsto KNO_3

\\ \large\sf\longmapsto 39u+14u+3(16u)

\\ \large\sf\longmapsto 53u+48u

\\ \large\sf\longmapsto 101u

\\ \large\sf\longmapsto 101g/mol

Now

\boxed{\sf No\:of\:moles=\dfrac{Given\:mass}{Molar\:mass}}

\\ \large\sf\longmapsto No\:of\:moles=\dfrac{0.565}{101}

\\ \large\sf\longmapsto No\:of\:moles=0.005mol

We know

\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Vol\:of\:Solution\:in\:L}}

\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{\dfrac{250}{1000}L}

\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{0.250}

\\ \large\sf\longmapsto Molarity=0.02M

8 0
3 years ago
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Hi pls answer this i don't get it ​
joja [24]
The answer is P
You can count how many electrons in the picture
Or you can search for the one who has 3 electrons in the outer shell
6 0
3 years ago
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