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MrRa [10]
3 years ago
12

How many grams are in 6.00 moles of NaCl?

Chemistry
2 answers:
VLD [36.1K]3 years ago
6 0

350.65662000000003 us your answer

Anastasy [175]3 years ago
5 0

Is it 350.4, I assume?

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7.2 mol NacN



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If the temperature of the helium balloon were increased from 30°C to 35°C and the volume of the balloon only expanded from 0.47L
Cerrena [4.2K]

Answer:r u from ridge

Explanation:

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4 0
3 years ago
A _____ solution contains more dissolved solute than a saturated solution at the same temperature.
klemol [59]
The answer according to my teacher would be Supersaturated
5 0
3 years ago
Which material does not melt at any temperature?
lord [1]

Answer: Option (c) is the correct answer.

Explanation:

Wood is a mixture of different substances. Primarily it consists of cellulose, lignin, water etc.

When we heat wood then all these substance oxidize into the atmosphere even before they could melt.

Whereas iron, sodium chloride and ethanol all are the substances which can melt at any temperature.

Thus, we can conclude that out of the given options, wood, a mixture of different substances is a material that does not melt at any temperature.

6 0
3 years ago
A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

8 0
3 years ago
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