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Makovka662 [10]
3 years ago
10

The empirical formula for carbohydrates is CH2O. The molecular weight for a particular carbohydrate is 90.09 grams. What is the

molecular formula for that carbohydrate? (atomic masses: C =12.0107; H = 1.00794; O = 15.9994)
Chemistry
1 answer:
baherus [9]3 years ago
6 0
We find the weight of the empirical formula:
12.0107 + 2 x 1.00794 + 15.9994
= 30.03

Now, we divide the molecular weight by the weight of the empirical formula to find the number of times the empirical formula repeats:
90.09 / 30.03
= 3

The formula is 3(CH₂O)
C₃H₆O₃
You might be interested in
A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the tempe
Vlad1618 [11]

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Explanation:

Given: V_{1} = 571 mL,       T_{1} = 26^{o}C

(a) T_{2} = 5^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL

(b) T_{2} = 95^{o}F

Convert degree Fahrenheit into degree Cesius as follows.

(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL

(c) T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL

8 0
2 years ago
How many grams of chlorine gas are present in a 150. liter cylinder of chlorine held at a pressure of 1.00 atm and 0. °C? Group
OlgaM077 [116]

Answer:

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

  • P= 1.00 atm
  • V= 150 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 0 C= 273 K

Replacing:

1.00 atm* 150 L= n*0.08206 \frac{atm*L}{mol*K} *273 K

Solving:

n=\frac{1.00 atm* 150 L}{0.08206 \frac{atm*L}{mol*K}*273 K}

n= 6.69 moles

Being Cl= 35.45 g/mole, the molar mass of chlorine gas is:

Cl₂=2*35.45 g/mole= 70.9 g/mole

So if 1 mole has 70.9 grams, 6.69 moles of the gas, how much mass does it have?

mass=\frac{6.69 moles*70.9 grams}{1 mole}

mass= 474.321 grams ≅ 474 grams

<u><em>474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C</em></u>

4 0
2 years ago
ANSWER FAST PLZ!!!!!!!!!
valina [46]
I think the answer is B
7 0
2 years ago
Read 2 more answers
Help with this question for chemistry!!
wel
Emperic formula is SO subscript 2. Molecular formula would be SO subscript 2 multiply 1 so it's the same answer.
7 0
2 years ago
The two main classes of mixtures are
Vesnalui [34]
Homogeneous or heterogeneous, Hope this helps!!
3 0
2 years ago
Read 2 more answers
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