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Natalka [10]
3 years ago
15

A man decides to lose weight by dieting and exercising. He chooses to reduce his caloric intake by 500 calories a day, and he li

fts weights daily as his main form of exercise. After strictly following his diet and exercise plan for two weeks, he finds that even though his waist has shrunk by two inches, he has actually gained weight. The man hypothesizes that his caloric intake is still too high. What might be another possible hypothesis?

Chemistry
1 answer:
Ivenika [448]3 years ago
8 0
Jdjdjdfbfbfbnfdjfjfjjf. Dd f f f rn

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How many moles are in 48.0g of H2O2?
Tju [1.3M]

Answer:

1.41 moles H2O2(with sig figs)

Explanation:

okay so what is the molar mass of H2O2= (1.008 g/mol)2+(16.00g/mol)2= (2.016+ 32.00) g/ mol

= 34. 02 g/mol

48.0g H2O2* 1 mol H2O2/ 34.02 g H2O2= 1.41 mol H2O2

3 0
3 years ago
Chemistry help! Thanks to anyone who helps!
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A: 2-methyl-2-propanom
5 0
3 years ago
How is oxygen different than neon explain your answer
lora16 [44]
Oxygen and neon are both elements. Oxygen has 8 electrons and 8 protons. Neon has 10 electrons and 10 protons. Oxygen is also a non-metal element and Neon is a noble gas.
5 0
3 years ago
2.The rate constant of a first order reaction is 66 s-1. What is the rate constant in units of minutes
Vanyuwa [196]

Answer:

66s^-1 will be 1/66

then to convert to minute you multiply by 69

1/66 x 60 = 3960 mins

4 0
3 years ago
The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t
slava [35]

<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol

The given chemical equation follows:

2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = \frac{1}{2}\times 0.133=0.0665moles of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

8 0
3 years ago
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