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LuckyWell [14K]
3 years ago
9

Write the formula for each of the following compounds: a. sulfur difluoride b. sulfur hexafluoride c. sodium dihydrogen phosphat

e d. lithium nitride e. chromium(III) carbonate f. tin(II) fluoride g. ammonium acetate h. ammonium hydrogen sulfate i. cobalt(III) nitrate j. mercury(I) chloride k. potassium chlorate l. sodium hydride
Chemistry
1 answer:
Bogdan [553]3 years ago
7 0

Answer:

a. sulfur difluoride SF₂

b. sulfur hexafluoride SF₆

c. sodium dihydrogen phosphate NaH₂PO₄

d. lithium nitride Li₃N

e. chromium(III) carbonate Cr₂(CO₃)₃

f. tin(II) fluoride SnF₂

g. ammonium acetate NH₄(CH₃COO)

h. ammonium hydrogen sulfate NH₄(HSO₄)

i. cobalt(III) nitrate Co(NO₃)₃

j. mercury(I) chloride Hg₂Cl₂

k. potassium chlorate KClO₃

l. sodium hydride NaH

Explanation:

The names give us information about the composition. First, we mention the cation  and then the anion. In the formula, we follow the same order. Each part has a charge but the resulting compound is electrically neutral.

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Which option is a key point of cell theory? Cells can repair neighboring cells, but not themselves.Cells can repair neighboring
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Answer:

Cells are the basic structural unit of all living organisms.

Explanation:

Hope this helps, but i'm unsure if it would be multiple choice or not.

3 0
2 years ago
Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 4.1 g of butane is m
m_a_m_a [10]

Answer:

12.44 g

Explanation:

2C4H10 + 13O2 = 8CO2 + 10H2O

n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).

n(O2) = m(O2)/M(O2) = 25.9 / 32g/mol = 0.809 mol (deficiency).

Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.

mass of CO2 produced =

M = [0.0707 moles C4H10 x 8 moles CO2] / 2 moles C4H10 x 44 g CO2/mol

M = 0.5656/2 * 44

M = 0.2828 * 44

M = 12.44 of CO2

5 0
3 years ago
A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a
Semenov [28]

<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

  • To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol

  • The chemical equation for the reaction of NaOH and sulfuric acid follows:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = \frac{1}{2}\times 0.01670=0.00835mol of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

  • Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

  • The chemical equation for the reaction of metal (forming M^{3+} ion) and sulfuric acid follows:

2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = \frac{2}{3}\times 0.0556=0.0371mol of metal

  • To calculate the molar mass of metal for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

6 0
3 years ago
Consider the balanced equation:
kakasveta [241]

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well hello there ms gabby I might not know the answer but you could text me on Ig or sc

ig:wavy_sean5

sc:despeechless5

ur really pretty

5 0
3 years ago
How many grams is 4.2X10^24 atoms of sulfur?
ipn [44]
First find the number of moles of sulfur using dimensional analysis with avogadro’s number as the conversion factor. 4.2*10^24 atoms * (1 mol/6.022*10^23 atoms) = 7.0 mol sulfur. The molar mass of sulfur is 32.06 g/mol, which is found on the periodic table as sulfur’s (S) atomic weight. Use dimensional analysis again with the molar mass of sulfur as the conversion factor. 7.0 mol * 32.06 g/mol = 224.42 g sulfur. Since the problems gives us two significant figures, round the mass of sulfur to 220 grams, or 2.2 * 10^2 g.
7 0
3 years ago
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