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Jlenok [28]
3 years ago
14

Determine whether or not each chemical formula is an empirical formula. C6H12O6 CaCO3 NaMnO4 Ba3(PO4)2 K2C2O4

Chemistry
1 answer:
zavuch27 [327]3 years ago
5 0

Answer:

- C6H12O6 is not an empirical formula because it can be simplified to CH2O.

- CaCO3 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- NaMnO4 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- Ba3(PO4)2 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- K2C2O4 is not an empirical formula because it can be simplified to KCO2.

Explanation:

Hello!

In this case, since the empirical formulas are referred to those formulas that have been simplified up to the smallest whole number for each subscript per atom, we can see that:

- C6H12O6 is not an empirical formula because it can be simplified to CH2O.

- CaCO3 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- NaMnO4 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- Ba3(PO4)2 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- K2C2O4 is not an empirical formula because it can be simplified to KCO2.

Best regards!

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A negatively charged balloon has 2.6 µc of charge. how many excess electrons are on this balloon? the elemental charge is 1.6 ×
Salsk061 [2.6K]
The given 2.6 µC of charge is due to a buildup of electrons, each of which has a charge of 1.6 x 10^-19 C. The 2.6 <span>µC is equivalent to 2.6 x 10^-6 C, so we can divide this by the individual charge of an electron:
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3 years ago
An experiment shows that a 236 mL gas sample has a mass of 0.443 g at a pressure of 740 mmHg and a temperature of 22 ∘C. What is
aleksley [76]

Answer:

49.2 g/mol

Explanation:

Let's first take account of what we have and convert them into the correct units.

Volume= 236 mL x (\frac{1 L}{1000 mL}) = .236 L

Pressure= 740 mm Hg x (\frac{1 atm}{760 mm Hg})= 0.97 atm

Temperature= 22C + 273= 295 K

mass= 0.443 g

Molar mass is in grams per mole, or MM= \frac{mass}{moles} or MM= \frac{m}{n}. They're all the same.

We have mass (0.443 g) we just need moles. We can find moles with the ideal gas constant PV=nRT. We want to solve for n, so we'll rearrange it to be

n=\frac{PV}{RT}, where R (constant)= 0.082 L atm mol-1 K-1

Let's plug in what we know.

n=\frac{(0.97 atm)(0.236 L)}{(0.082)(295K)}

n= 0.009 mol

Let's look back at MM= \frac{m}{n} and plug in what we know.

MM= \frac{0.443 g}{0.009 mol}

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3 0
3 years ago
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Answer:

Explanation:

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e) 1.000 g X: 0.2733 g Y

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7 0
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