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3 years ago

Describe what atoms are.

Chemistry

1 answer:

umka21 [38]3 years ago

5 0

Answer:

the smallest component of an element having the chemical properties of the element

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Marina CMI [18]

Answer:

Explanation:

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2 years ago

what mass of carbon dioxide gas would be produced if 10g of calcium carbonate reacted with an excess of hydrochloric acid?

LuckyWell [14K]

Answer is: 4,4 grams <span>of carbon dioxide gas would be produced.
</span>Chemical reaction: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O.
m(CaCO₃) = 10 g.
n(CaCO₃) = 10 g ÷ 100 g/mol.
n(CaCO₃) = 0,1 mol.
From chemical reaction: n(CaCO₃) : n(CO₂) = 1 : 1.
n(CO₂) = 0,1 mol.
m(CO₂) = n(CO₂) · M(CO₂).
m(CO₂) = 0,1 mol· 44 g/mol.
m(CO₂) = 4,4 g.

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3 years ago

Does respiratory happens in the lungs true or false?

Anna007 [38]

Answer:

True

Explanation:

7 0

2 years ago

Read 2 more answers

Determine the number of moles in 258.45 grams of iron (III) chloride (FeCl3). Show and explain all

QveST [7]

Answer:

$\boxed {\boxed {\sf 1.5935 \ mol \ FeCl_}}$

Explanation:

To convert from grams to moles, we must use the molar mass. This can be found on the Periodic Table. First, find the molar mass of iron and chlorine.

Fe: 55.84 g/mol
Cl: 35.45 g/mol

Check the formula. There is a subscript of 3 after Cl, so there are 3 atoms of chlorine in 1 molecule. Multiply iron's molar mass by 3, then add iron's molar mass.

FeCl₃: 55.84 + 3(35.45) = 55.84+106.35=162.19 g/mol

Use this number as a ratio.

$\frac {162.19 \ g\ FeCl_3}{1 \ mol \ FeCl_3}$

Multiply by the given number of grams.

$258.45 \ g \ FeCl_3 *\frac {162.19 \ g\ FeCl_3}{1 \ mol \ FeCl_3}$

Flip the ratio so the grams of iron (III) chloride cancel.

$258.45 \ g \ FeCl_3 *\frac {1 \ mol \ FeCl_3}{162.19 \ g\ FeCl_3}$

$258.45 *\frac {1 \ mol \ FeCl_3}{162.19}$

$\frac {258.45 \ mol \ FeCl_3}{162.19}$

$1.59350144892 \ mol \ FeCl_3$

The original measurement of grams has 5 significant figures, so our answer must have the same. For the number we calculated, that is the ten thousandth place.

$1.5935 \ mol \ FeCl_3$