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Svet_ta [14]
2 years ago
12

What is the molar mass of NaC2H302?

Chemistry
1 answer:
andrezito [222]2 years ago
3 0

Answer:

82.0343 g/mol

Explanation:

Count each element and the number of atoms for each element. Add them all together. Use the periodic table.

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1.00 g of a compound is combusted in oxygen and found to give 3.14g of CO2 and 1.29 g of H2O. From these data we can tell thatA.
lora16 [44]

Answer:

the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula

Explanation:

Mass of CO2 obtained = 3.14 g

Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol

The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g

Mass of H2O obtained = 1.29 g

Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol

The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g

% by mass of carbon = 0.857/1 ×100 = 85.7 %

% by mass of hydrogen = 0.0717/1 × 100 = 7.17%

Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %

Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.

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3 years ago
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alexgriva [62]

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Explanation:

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2 years ago
How many valence electrons are lost by Sr when forming SrI2?
Ostrovityanka [42]
Two
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How many moles are in 297g of nh3?<br><br> Please show work, will give brainliest.
rusak2 [61]

Answer:

17.5moles

Explanation:

The number of moles in a substance can be calculated by using the formula;

Number of moles (n) = mass (m) ÷ molar mass (MM)

According to this question, mass of ammonia (NH3) = 297g

Molar Mass of NH3 = 14 + 1(3)

= 17g/mol

n = 297/17

n = 17.47

Number of moles of NH3 = 17.5moles

3 0
3 years ago
A hypothetical metal crystallizes with the face-centered cubic unit cell. The radius of the metal atom is 234 picometers and its
AleksAgata [21]

Answer:

\delta=101.13 g/cm^3

Explanation:

As can be seen in the Figure in a face-centered cubic unit cell you have:

  • Six halves of atoms
  • Eight 1/8 of atom (1 in each corner)

In total:

n_{atom}=6*0.5+8*\frac{1}{8}=4 atoms

Now, each side of the cell is 234 picometers (2.34e-8 cm) long

V_{cell}=L^3=(2.34e^{-8} cm)^3

V_{cell}=1.28*10^{-23}cm^3

Atoms per cm3:

n=\frac{4 atoms}{1.28*10^{-23}cm^3}

n=3.12*10^{23} atoms/cm^3

Expressing in mass:

\delta=3.12*10^{23} atoms/cm^3* \frac{1 mol}{6.022*10^{23}}*195.08 g/mol

\delta=101.13 g/cm^3

6 0
3 years ago
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