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Advocard [28]
2 years ago
14

In witch step did a chemical change most likely occur

Chemistry
1 answer:
11Alexandr11 [23.1K]2 years ago
6 0

Answer:

step 2

hope this helps!

please mark as brainiest<3

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The balanced equation for water is 2 H2 + O2 to 2 H2O. If I have 21.2g of a product , and I started with 5.6 g of H2, how many g
ANTONII [103]

Since 21.2 g H2O was produced, the amount of oxygen that reacted can be obtained using stoichiometry. The balanced equation was given:  2H₂ + O₂ → 2H₂O and the molar masses of the relevant species are also listed below. Thus, the following equation is used to determine the amount of oxygen consumed.

Molar mass of H2O = 18 g/mol

Molar mass of O2 = 32 g/mol

21.2 g H20 x 1 mol H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 18.8444 g O2

<span>We then determine that 18.84 g of O2 reacted to form 21.2 g H2O based on stoichiometry. It is important to note that we do not need to consider the amount of H2 since we can derive the amount of O2 from the product. Additionally, the amount of H2 is in excess in the reaction.</span>

7 0
3 years ago
A neutron collides with a nitrogen atom, resulting in a transmutation. Balance the equation. Superscript 1 Subscript 0 Baseline
o-na [289]

Answer:

A is 14

Z is 6

X is C

Explanation:

Have a great day!

7 0
2 years ago
Sometimes called the "River of Grass," _________ Is a vast wetland
timama [110]

Answer: A

Explanation:

i just know

6 0
2 years ago
Read 2 more answers
Bacteria are very small *<br><br> True<br> False
elena-14-01-66 [18.8K]

Of course they are small

Explanation:

The only way you can see them is by a microscope or a lens and can be anywhere.

8 0
3 years ago
How much heat is required to heat 1.6g of ice from -16c to steam at 112c?
oksian1 [2.3K]

Answer:

Total heat ≅ 49.07 kJ

Explanation:

Given that:

mass = 1.6 g = 0.016 kg

Initial temperature = - 16 ° C

final temperature = 112° C

specific heat for ice = 2.06 kJ/kgC

specific heat of water = 4.186 kJ/kgC

heat fusion of ice = 334 kJ/kg

specific heat for steam = 2.1 kJ/kgK

heat of vaporization of water = 2256 kJ/kg

To heat ice from -16 ° C to 0 ° C

Q₁ =  2.06 kJ/kgC × 0.016 kg ×  16 ° C

Q₁ =  0.52736 kJ

To melt Ice at 0° C

Q₂= 334 kJ/kg × 0.016 kg = 5.344 kJ

To heat water from 0° C to  100° C

Q₃ = 4.186 kJ/kgC × 0.016 kg  × 100° C

Q₃ = 6.6976 kJ

To vaporize water to steam at 100° C

Q₄ = 2256 kJ/kg × 0.016 kg = 36.096 kJ

To heat steam from 100C to 112° C

Q₅ = 2.1 kJ/kgC × 0.016 kg × 12 C

Q₅ = 0.4032 kJ

Total heat = Q₁ + Q₂ + Q₃ + Q₄  + Q₅

Total heat =  (0.52736 +  5.344 +  6.6976 + 36.096 + 0.4032) kJ

Total heat = 49.06816  kJ

Total heat ≅ 49.07 kJ

6 0
3 years ago
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