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Maurinko [17]
1 year ago
14

I) Choose The Correct Answer 1. Which of the following statement correctly describes the law of conservation of mass? A/ It stat

es that the total mass of reactants is equals with the total mass of products in all chemical reaction B/ The total numbers of atoms at the left should be equal with the total numbers of atoms at the right side of arrow C/Atom is neither created nor destroyed in any chemical reaction D/ All E/ None​
Chemistry
1 answer:
spin [16.1K]1 year ago
6 0

Answer:

<em><u>D</u></em><em><u> </u></em>all

Explanation:

because law of conservation of mass describes A, B and C statement

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Flammabilty. Is a chemical property that tells whether a substance reacts in the presence of carbon dioxide try or false
beks73 [17]

Answer:

False

Explanation:

Flammability is a chemical property that tells whether a substance reacts in the presence of extreme heat, spark, or another form of ignition.

4 0
3 years ago
Read 2 more answers
How many molecules are found in 90 grams of Al(OH)3?
jarptica [38.1K]

Answer:

6.95 x 10²³ molecules/particles

Explanation:

First we need to find the total Empirical Mass. We can find this by adding each element's mass together.

Al = 26.982,

O = 15.999

H = 1.008

26.982 + 3(15.999) + 3(1.008) = 78.003.

Now we divide by the mass given (90 grams).

90/78.003 = 1.153801777.

We then take that number and multiply it by avogadro's number (6.022 x 10²³)

1.153801777 x avogadro's number = 6.95 x 10²³

6 0
3 years ago
POINTSSS!!!!!!
Lelu [443]
1 mole CO₂----------- 6.02x10²³ molecules
4 moles  CO₂ ---------- ??

4 x ( 6.02x10²³) / 1 =

= 2.41 x 10²⁴ / 1 => 2.41 x 10²⁴ molecules of CO₂

Answer C
4 0
3 years ago
Use the unbalanced equation NH3 + O2 = NO + H2O
Elodia [21]
<span>The balance format is
4NH3+ 5O2 -------> 4NO + 6H2O </span>
7 0
3 years ago
The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd
kenny6666 [7]

Answer:

The answer is "= 0.078 \ kg \ H_2".

Explanation:

calculating the moles in CH_4 =\frac{PV}{RT}

                                                =\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol

Eqution:

CH_4 +H_2O \to  3H_2+ CO \ (g)

Calculating the amount of H_2 produced:

= 12.9 \ mol CH_4 \times  \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2

So, the amount of dihydrogen produced = 0.078 \frac{kg}{s}

5 0
3 years ago
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