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3 years ago

How do I set this up?

Chemistry

1 answer:

avanturin [10]3 years ago

8 0

Start with 2.95 g of aspartame. Find g per mole of aspartame. Then find moles of nitrogen. 2 N to 1 aspartame. Then there are 14.00067 g of N per 1 mole. Change g to milligrams with the 10^-3 g

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If a radioactive material has a 10 year half-life, how much of a 100 g sample will be left after 30 years?

olga_2 [115]

12.5g, each 10 years you lose a half of what you have at that given moment

5 0

3 years ago

A 3.0 g sample of a gas occupies a volume of 1.00L at 100C and 740 torr pressure. The molecular weight of the

SOVA2 [1]

Answer:

94.2 g/mol

Explanation:

Ideal Gases Law can useful to solve this

P . V = n . R . T

We need to make some conversions

740 Torr . 1 atm/ 760 Torr = 0.974 atm

100°C + 273 = 373K

Let's replace the values

0.974 atm . 1 L = n . 0.082 L.atm/ mol.K . 373K

n will determine the number of moles

(0.974 atm . 1 L) / (0.082 L.atm/ mol.K . 373K)

n = 0.032 moles

This amount is the weigh for 3 g of gas. How many grams does 1 mol weighs?

Molecular weight → g/mol → 3 g/0.032 moles = 94.2 g/mol

3 0

3 years ago

Sound is energy moving in

galben [10]

Whats do you want to know?

7 0

3 years ago

2.What two factors must be held constant for density to be a constant ratio?

irina1246 [14]

Answer:

Temperature and Pressure

Explanation:

Temperature and pressure cause change in volume.

So any change in volume will alter the ratio of density as given by equation of density.

Density = mass/ volume

Change in volume will alter the ratio.

Kindly mark it branliest if the answer is little bit satisfying.

6 0

3 years ago

Read 2 more answers

When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H

algol [13]

Answer: The empirical formula is $CH_2$ and molecular formula is $C_4H_8$

Explanation:

We are given:

Mass of $CO_2$ = 18.95 g

Mass of $H_2O$ = 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, = $\frac{12}{44}\times 18.59=5.07g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, = $\frac{2}{18}\times 7.759=0.862g$ of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.422}{0.422}=1$

For H = $\frac{0.862}{0.422}=2$

The ratio of C : H = 1: 2

Hence the empirical formula is $CH_2$ .

The empirical weight of $CH_2$ = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4$

The molecular formula will be= $4\times CH_2=C_4H_8$

6 0

3 years ago