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STALIN [3.7K]
2 years ago
6

Hydrogen gas reacts with oxygen gas to form water. What is the excess reactant in this recation(Refer to the picture below to an

swer).
A. Water
B. Hydrogen gas
C. Oxygen gas
D. Water and oxygen gas

Chemistry
1 answer:
Studentka2010 [4]2 years ago
7 0

Answer: C. Oxygen Gas

Explanation: When Hydrogen Gas and Oxygen Gas combine together it forms water correct. When you have an excess amount of another it stays neutral because it has no other molecule to react with.

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How moles are in 45.7g of CH3?
miskamm [114]

first find the atomic weight of CH3 which would be

atomic weight: 12.011 (3×1.008) = 36.32 g/mol

then find the moles in the given mass

36.32 ÷ 45.7 = 0.794

I HOPE I'M NOT WRONG I HAVENT DONE CHEM IN SO LONG

6 0
3 years ago
Na2o + h2so4 ———&gt; na2so4 + h2o<br><br><br> Can anyone balance this please!!??
marysya [2.9K]

Answer:

The equation is already balanced. There's an equal number of materials on each side of the equation.

4 0
3 years ago
the rate of disappearance of Br- at some moment in time was determined to be 3.5 x 10-4 M/s. What is the rate of appearance of B
ddd [48]

Answer:

1.8 × 10⁻⁴ mol M/s

Explanation:

Step 1: Write the balanced reaction

2 Br⁻ ⇒ Br₂

Step 2: Establish the appropriate molar ratio

The molar ratio of Br⁻ to Br₂ is 2:1.

Step 3: Calculate the rate of appearance of Br₂

The rate of disappearance of Br⁻ at some moment in time was determined to be 3.5 × 10⁻⁴ M/s. The rate of appearance of Br₂ is:

3.5 × 10⁻⁴ mol Br⁻/L.s × (1 mol Br₂/2 mol Br⁻) = 1.8 × 10⁻⁴ mol Br₂/L.s

3 0
2 years ago
Convert 9.78 gallons toliters
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Answer:

44.46076 litres

Explanation:

5 0
3 years ago
Read 2 more answers
Use the reaction given below to solve the problem that follows: Calculate the mass in grams of aluminum oxide produced by the re
bearhunter [10]

Answer:  28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}   

\text{Moles of} Al=\frac{15.0g}{27g/mol}=0.556moles

The balanced chemical equuation is:

4Al+3O_2\rightarrow 2Al_2O_3  

According to stoichiometry :

4 moles of Al produce == 2 moles of Al_2O_3

Thus 0.556 moles of Al will produce=\frac{2}{4}\times 0.556=0.278moles  of Al_2O_3

Mass of Al_2O_3=moles\times {\text {Molar mass}}=0.278moles\times 102g/mol=28.4g

Thus 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal.

7 0
2 years ago
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