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2. Calculate the mass of solvent in grams in a solution containing 3.0 grams of Tylenolif the mass percent is 3.5%.

Answers

Answer 1

Answer:

Answer:

83g

Explanation:

1, Formula

Mass percent = (mass of solute/mass of solution) × 100

2. Determine mass of solution

Substitute the data and clear the mass of solution:

3.5 = (3.0g / mass of solution) × 100

0.035 = 3.0g / mass of solution

mass of solution = 3.0 g / 0.035

mass of solution = 85.714g

3. Determine the mass of solvent:

mass of solvent = mass of solution - mass of solute

mass of solvent = 85.714g - 3.0g = 82.714g

Round to two significant figures: 83 g

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Answers

D volcanic corruption

James adds some magnetic marbles to a glass jar full of ordinary marbles, and then shakes up the jar.

Answers

Answer: Magnetic marbles will tend to attract each other

50.0ml each of 1.0M Hcl and 1.0M Naoh at room temperature (20.0c) are mixed the temperature of the resulting Nacl solutions increase to 27.5cthe density if the resulting Nacl solutuion 1.02 g/ml
the specific heat of the resulting Nacl solutions is 4.06j/gc
calculate the heat of neutralisation of hcl and naoh in kj/mol nacl products

Answers

Answer:

62.12kJ/mol

Explanation:

The neutralization reaction of HCl and NaOH is:

HCl + NaOH → NaCl + H₂O + HEAT

You can find the released heat of the reaction and heat of neutralization (Released heat per mole of reaction) using the formula:

Q = C×m×ΔT

Where Q is heat, C specific heat of the solution (4.06J/gºC), m its mass and ΔT change in temperature (27.5ºC-20.0ºC = 7.5ºC).

The mass of the solution can be finded with the volume of the solution (50.0mL of HCl solution + 50.0mL of NaOH solution = 100.0mL) and its density (1.02g/mL), thus:

100.0mL × (1.02g / mL) = 102g of solution.

Replacing, heat produced in the reaction was:

Q = C×m×ΔT

Q = 4.06J/gºC×102g×7.5ºC

Q = 3106J = 3.106kJ of heat are released.

There are 50.0mL ×1M = 50.0mmoles = 0.0500 moles of HCl and NaOH that are reacting releasing 3.106kJ of heat. That means heat of neutralization is:

3.106kJ / 0.0500mol of reaction =

62.12kJ/mol is heat of neutralization

What is the total mass (amu) of carbon in each of the following molecules?(a)CH4
(b)CHCL3
(c)C12H10O16
(d)CH3CH2CH2CH2CH3

Answers

The mass of carbon in $\rm CH_4$ is 12.007 amu, $\rm CHCl_3$ is 12.007, $\rm C_1_2H_1_0O_1_6$ is 144.084 amu, and $\rm CH_3CH_2CH_2CH_2CH_3$ is 60.035 amu.

What is the mass of one Carbon atom?

The mass has been given as the sum of the atomic mass unit in the compound. The mass of 1 atom of carbon is 12.007 amu.

The mass of carbon in the following compounds is given as:

$\rm CH_4$

The number of Carbon units = 1

The mass of carbon in compound = 12.007 amu

$\rm CHCl_3$

The number of Carbon units = 1

The mass of carbon in the compound = 12.007 amu

$\rm C_1_2H_1_0O_1_6$

The number of carbon units =12

The mass of carbon in the compound:

$\rm 12\;*\;12.007\;amu\n=144.084\;amu$

$\rm CH_3CH_2CH_2CH_2CH_3$

The number of carbon units = 5

The mass of carbon in the compound:

$\rm 5\;*\;12.007\;amu\n=60.035\;amu$

Learn more about the mass of an atom, here:

brainly.com/question/5566317

Answer:

The atomic mass of carbon (C) is 12.0107 amu, so if you want to calculate the total mass in each molecule, you just need to multiply the number of carbon atoms in the substance by 12.017. In (a) there is one atom of C, (b) have also one atom of C, (c) have 12 atoms of C, and (d) have five atoms of C. Thus, the total mass (amu) of carbon is:

(a) 12.017 amu

(b) 12.017 amu

(d) 60.085 amu

Ammonia is produced by the reaction of nitrogen and hydrogen: N2(g) + H2(g)  NH3(g)(a) Balance the chemical equation.
(b) Calculate the mass of ammonia produced when 35.0g of nitrogen reacts with hydrogen.

Answers

Answer:

a) N2 (g) + H2 = 2 NH3

b) You have to state the mass of hydrogen

If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is

Answers

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a

compound.

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

Mn O

% composition 72.1 27.9

Divide by mass number 54.94 16

1.31 1.74

Divide by the smallest number 1.31 1.31

1 1.3

The resulting ratio is 1:1

Hence the Empirical formula is MnO, Manganese oxide

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