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Tasya [4]
3 years ago
5

A _______ pattern organizes a speech by incorporating repetition and variations of themes and ideas

Chemistry
1 answer:
astraxan [27]3 years ago
8 0

Answer:

A wave pattern organizes a speech

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Group 18 is known as the Noble/ Inert Gases
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Which of the following is true about incomplete dominance? Options: 1. Neither allele is completely dominant over the other alle
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<span>3. The parents’ phenotypes are expressed equally in the offspring’s phenotype.</span>
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2Fe(OH)3 Fe2O3+3H2O how many grams of Fe2O3 are produced if 10.7 grams of Fe (OH)3 react in this way
sasho [114]

Answer: 7.98 grams of Fe_2O_3 are produced if 10.7 grams of Fe(OH)_3 are reacted.

Explanation:

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

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

Putting values in equation 1, we get:

\text{Moles of} Fe(OH)_3=\frac{10.7g}{106.87g/mol}=0.100mol  

The chemical equation for the reaction is

2Fe(OH)_3\rightarrow Fe_2O_3+3H_2O

By Stoichiometry of the reaction:

2 moles of Fe(OH)_3 produce = 1 mole of Fe_2O_3

So, 0.100 moles of Fe(OH)_3 produce= \frac{1}{2}\times 0.100=0.05mol of Fe_2O_3  

Mass of Fe_2O_3 =moles\times {\text{Molar Mass}}=0.05mol\times 159.69g/mol=7.98g  

Hence 7.98 grams of Fe_2O_3 are produced if 10.7 grams of Fe(OH)_3 are reacted.

3 0
3 years ago
Write the correct chemical formula for potassium and sulfur
Shalnov [3]
Phosphorus + Sulfur ------> Phosphorus sulfide

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Hope it helped!
3 0
3 years ago
What volume of 0.235 M H2SO4 is needed to titrate 40.0 mL of 0.0500 M Na2CO M NaCO3 solution?
Andreyy89

Answer: The volume of 0.235 M H_2SO_4 needed to titrate 40.0 mL of 0.0500 M Na_2CO_3 is 8.51 ml

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Na_2CO_3

We are given:

n_1=2\\M_1=0.235M\\V_1=?mL\\n_2=2\\M_2=0.0500M\\V_2=40.0mL

Putting values in above equation, we get:

2\times 0.235\times V_1=2\times 0.0500\times 40.0\\\\V_1=8.51mL

Thus volume of 0.235 M H_2SO_4 needed to titrate 40.0 mL of 0.0500 M Na_2CO_3 is 8.51 ml

5 0
3 years ago
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