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Natali [406]
3 years ago
5

A solid with a density of 3.57g/mL and mass of 19.5g is added to a graduated cylinder that contain 23.2mL of water. What is the

volume of the liquid once the solid is dropped in the cylinder?
Chemistry
1 answer:
NNADVOKAT [17]3 years ago
3 0

Answer:

Volume of liquid = 28.7 mL

Explanation:

Given data;

Density of solid = 3.57 g/ml

Mass of solid = 19.5 g

Volume of water = 23.2 mL

Total volume when solid is dropped into graduated cylinder= ?

Solution:

Density = mass/ volume

v = m/d

v = 19.5 g/ 3.57 g/ml

v = 5.5 mL

Volume of liquid = volume of water + volume of solid

Volume of liquid = 23.2 mL + 5.5 mL

Volume of liquid = 28.7 mL

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2. Would you measure a pencil in meters? A hallway in millimeters? Discuss
Eduardwww [97]

Answer:

Measuring a pencil in meters would be very difficult, as a single meter is much longer than one pencil. Also, measuring a hallway in millimeters would be very difficult considering how small millimeters are in comparison to a hallway. However, if you switch these two then they would work very well.

5 0
3 years ago
3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15
shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}

(b) Mass of H

\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

\text{Moles of C = 1400  mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016  + 16.00) u  = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00  \approx 5

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0
3 years ago
I need help with homework
stellarik [79]

The equation that conforms with the particle model is 4C + 6O2 ----> 4CO2 + 2O2.

<h3>What is a chemical reaction?</h3>

The term chemical reaction has to do with the combination of substances to give products. We know that reactants and products must follow the law of conservation of mass.

Hence, the equation that conforms with the particle model is 4C + 6O2 ----> 4CO2 + 2O2.

Learn more about chemical reaction:brainly.com/question/22817140

#SPJ1

4 0
2 years ago
Dalton's law of partial pressures states that the total pressure exerted by a mixture of gases is the sum of the pressures exert
Damm [24]

Answer: The given statement is true.

Explanation:

According to the Dalton's law, total pressure of a mixture of gases that do not react with each other is equal to the partial pressure exerted by each gas.

The relationship is as follows.

          p_{total} = \sum_{i=1}^{n} p_{i}

or,        p_{total} = p_{1} + p_{2} + p_{3} + p_{4} + ......... + p_{n}

where,  p_{1}, p_{2}, p_{3} ....... = partial pressure of individual gases present in the mixture

Also, relation between partial pressure and mole fraction is as follows.

                 p_{i} = p_{total} \times x_{i}

where,      x_{i} = mole fraction

Thus, we can conclude that the statement Dalton's law of partial pressures states that the total pressure exerted by a mixture of gases is the sum of the pressures exerted independently by each gas in the mixture, is true.              

5 0
3 years ago
What is the ph of a solution of 0.550 m k2hpo4, potassium hydrogen phosphate?
Solnce55 [7]
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+  + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:

PH= -㏒[H+]
     = -㏒(4.81x10^-7) = 6.32

3 0
3 years ago
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