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2 years ago

A 26.0-g sample of a liquid was found to have a volume of 13.0 ml. what is the density of the liquid

Chemistry

2 answers:

Bond [772]2 years ago

7 0

The density of a substance is simply the ratio of mass over volume. The formula for density is therefore:

density = mass / volume

Substituting the values:

density = 26 g / 13 mL

<span>density = 2 g / mL</span>

Komok [63]2 years ago

3 0

<u>Answer:</u> The density of liquid is 2 g/mL

<u>Explanation:</u>

To calculate density of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

We are given:

Mass of liquid = 26.0 g

Volume of liquid = 13.0 mL

Putting values in above equation, we get:

$\text{Density of liquid}=\frac{26.0g}{13.0mL}=2g/mL$

Hence, the density of liquid is 2 g/mL

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What is the percent yield of this reaction if 22.o g of Mgl2 is produced by the reaction of 25.0 g of Mg with 25.0 g of l2?

expeople1 [14]

% yield = 80.719

<h3>Further explanation</h3>

Given

22.0 g of Mgl₂

25.0 g of Mg

25.0 g of l₂

Required

The percent yield

Solution

Reaction

Mg + I₂⇒ MgI₂

mol Mg = 25 g : 24.305 g/mol = 1.029

mol I₂ = 25 g : 253.809 g/mol = 0.098

Limiting reactant = I₂

Excess reactant = Mg

mol MgI₂ based on I₂, so mol MgI₂ = 0.098

Mass MgI₂ (theoretical):

= mol x MW

= 0.098 x 278.114

= 27.255 g

% yield = (actual/theoretical) x 100%

% yield = (22 / 27.255) x 100%

% yield = 80.719

8 0

1 year ago

Suppose you are studying the K sp of K C l O 3 , which has a molar mass of 122.5 g/mol, at multiple temperatures. You dissolve 4

Ghella [55]

Answer:

The K sp Value is $K_{sp}=7.40$

Explanation:

From the question we are told that

The of $KClO_3$ is = 122.5 g/ mol

The mass of $KClO_3$ dissolved is $m_s = 4.0g$

The volume of solution is $V_s = 12mL = 12*10^{-3}L$

The number of moles of $KClO_3$ is mathematically evaluated as

$No \ of \ moles \ = \frac{mass }{Molar \ mass}$

Substituting values

$= \frac{4}{122.5}$

$=0.0327\ moles$

Generally concentration is mathematically represented as

$concentration = \frac{No \ of \ moles}{volume }$

For $KClO_3$

$Z= \frac{0.0327}{12*10^{-3}}$

$=2.72 \ mol/L$

The dissociation reaction of $KClO_3$ is

$KClO_3 \ ----> K^{+}_{(aq)} + ClO_3^-_{(aq)}$

The solubility product constant is mathematically represented as

$K_{sp} = \frac{concentration of ionic product }{concentration of ionic reactant }$

Since there is no ionic reactant we have

$K_{sp} = [k^+] [ClO_3^-]$

$= Z^2$

$= 2.72^2$

$K_{sp}=7.40$

5 0

2 years ago

Read 2 more answers

How many grams are in 2.5 pound sample

julsineya [31]

About 2,500 grams Ans balkfdoaks;

5 0

2 years ago

If 4.59 g of potassium reacts with 3.6 g of sulfur according to the following reaction, how many grams of potassium sulfide can

EleoNora [17]

First convert the amount of grams you have of each substance to moles. Find your limiting reactant by calculating how many grams are needed to complete this reaction. If done correctly, you would see that we need .226 moles of Potassium to complete this reaction. However, we only have .118 moles of Potassium, so K must be our limiting reactant. Then use the moles of K to find out how many moles of K^2S are made. Then convert the amount of moles of K^2S to grams and you should get 10.3 g K^2S

8 0

2 years ago

If one starts with pure NO2(g) at a pressure of 0.500 atm, the total pressure inside the reaction vessel when 2NO2(g) 2NO(g) + O

natulia [17]

Answer:

The partial pressure of NO2 = 0.152 atm

Explanation:

Step 1: Data given

Pressure NO2 = 0.500 atm

Total pressure at equilibrium = 0.674 atm

Step 2: The balanced equation

2NO2(g) → 2NO(g) + O2(g)

Step 3: The initial pressure

pNO2 = 0.500 atm

pNO = 0 atm

p O2 = 0 atm

Step 4: Calculate pressure at the equilibrium

For 2 moles NO2 we'll have 2 moles NO and 1 mol O2

pNO2 = 0.500 - 2x atm

pNO =2x atm

pO2 = xatm

The total pressure = p(total) = p(NO2) + p(NO) + p(O2)

p(total) = (0.500 - 2x) + 2x + x= 0.674 atm

0.500 + x = 0.674 atm

x = 0.174 atm

This means the partial pressure of NO2 = 0.500 - 2*0.174 = 0.152 atm

6 0

2 years ago