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notka56 [123]
4 years ago
14

In this chemical reaction, how many grams of h2 will react completely with6.50 moles of CO

Chemistry
1 answer:
geniusboy [140]4 years ago
3 0

52 g of hydrogen H₂

Explanation:

I will assume that the problem is talking about the hydrogenation of carbon dioxide CO₂ not carbon monoxide CO. It is harder to reduce carbon dioxide than carbon monoxide and if you manage to reduce carbon dioxide you can reduce the carbon monoxide as well.

This reaction it will take place in the presence of catalyst at a specific temperature and pressure.

CO₂ + 4 H₂ → CH₄ + 2 H₂O

Now taking into the account the chemical reaction we devise the following reasoning:

if         1 mole of CO₂ react with 4 moles of H₂

then   6.5 moles of CO₂ react with X moles of H₂

X = (6.5 × 4) / 1 = 26 moles of H₂

number of moles = mass / molecular wight

mass = number of moles × molecular wight

mass of H₂ = 26 × 2 = 52 g

Learn more about:

hydrogenation reaction

brainly.com/question/13592967

#learnwithBrainly

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A sound wave is traveling at 331 m/s and has a frequency of 126 Hz. What is the wavelength?​
damaskus [11]
<h2>Answer</h2>

2.626984127 m

<h2>Explanation:</h2><h2></h2>

You have to know the equation that relates wavelength, frequency, and velocity (it's like speed but a bit different).

v = f x λ

where:

v = velocity

f = frequency

λ = Wavelength

Rearrange to make λ subject:

λ = v / f

We've been given 331 as the speed, 126 as the frequency. Sub it into the equation:

331 / 126 = 2.626984127 m

3 0
3 years ago
How many moles of aluminum are needed to make 9 moles of molecular hydrogen? given the reaction: 2 al + 6 hcl → 2 alcl3 + 3h2 6
sergiy2304 [10]
1) Chemical equation

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2) molar ratios

2 mol Al : 3 moles H2

3) Proportion

2 mol Al /  3mol H2 = x / 9 mol H2

4) Solve for x

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7 0
3 years ago
Chemistry gcse
mart [117]
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8 0
3 years ago
Yet a third pair of compounds of manganese and oxygen is 50.48% and 36.81% oxygen respectively. In what small whole number ratio
Mariulka [41]

Answer:

The number ratio is 4:7

Explanation:

Step 1: Data given

Compound 1 has 50.48 % oxygen

Compound 2 has 36.81 % oxygen

Molar mass oxygen = 16 g/mol

Molar mass manganese = 54.94 g/mol

Step 2: Calculate % manganes

Compound 1: 100 - 50.48 = 49.52 %

Compound 2: 100 - 36.81 = 63.19 %

Step 3: Calculate mass

Suppose mass of compounds = 100 grams

Compound 1:

 50.48 % O = 50.48 grams

 49.52 % Mn = 49.52 grams

Compound 2:

36.81 % O = 36.81 grams

63.19 % Mn = 63.19 grams

Step 4: Calculate moles

Compound 1

Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles

Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles

Compound 2

Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles

Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles

Step 5: calculate mol ratio

We will divide by the smallest amount of moles

Compound 1

O: 3.155/0.9013 = 3.5

Mn: 0.9013 / 0.9013 = 1

Mn2O7

Compound 2

O: 2.301 / 1.150 = 2

Mn: 1.150 / 1.150 = 1

MnO2

The number ratio is 2:3.5 or 4:7

7 0
3 years ago
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