CHEMISTRY HIGH SCHOOL

26.0 g of a liquid that has a density of 1.44 g/mL needs to be measured out in a graduated cylinder. what volume should be used

Answers

Answer 1

Answer:

Answer: The volume of liquid used will be 18.055 mL.

Explanation:

To calculate the volume of liquid, we use the equation for density which is:

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

We are given:

Mass of the liquid = 26 g

Density of the liquid = 1.44 g/mL

Volume of the liquid = ? mL

Putting values in above equation, we get:

$1.44g/mL=\frac{26g}{\text{Volume of the liquid}}\n\n\text{Volume of the liquid}=18.055mL$

Hence, the volume of liquid used will be 18.055 mL.

Answer 2

Answer: As Density = Mass/Volume
Mass = 26.0g
Density = 1.44g/mL
Therefore Volume = Mass/Density
=> Volume = 26.0/1.44 = 18.055... = 18.1mL (to 3 sig figs)

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A copper penny has a mass of 6.2 grams and a volume of 0.70ml. What is the density of the penny

Answers

Given:
A copper penny has a mass of 6.2 grams and a volume of 0.70ml

Required:
density of the penny

Solution:
D = M/V where D is teh density, M is the mass and V is the volume

D = 6.2 grams/0.70 mL
D = 8.9 grams per milliliters

True or false? The radioactive wastes produced by nuclear fission remain dnagerous for dozens of years

Answers

True, this is why Chernobyl is still not to be lived in

If 20 milliliters of 4.0 M NaOH is exactly neutralized by 20 milliliters of HCL, the molarity of the HCL is

Answers

Volume of NaOH = 20 mL / 1000 = 0.02 L

Molarity NaOH = 4.0 M

number of moles NaOH :

n = M * V

n = 4.0 * 0.02

n = 0.08 moles of NaOH:

Finally we calculate the number of moles of HCl in the solution from the stoichiometry of the reaction :
HCl + NaOH = NaCl + H₂O

1 mole HCl ---------- 1 mole NaOH
? moles HCl --------- 0.08 moles NaOH

moles HCl = 0.08 * 1 / 1

= 0.08 moles of HCl

Volume of HCl = 20 mL / 1000 = 0.02 L

M ( HCl ) = n / V

M = 0.08 / 0.02

= 4.0 M

hope this helps!

Answer:

4.0

Explanation:

The particles that are lost, gained, or shared in chemical reactions are

Answers

Electrons are either lost gained or shared in chemical reactions :)

Ammonia and sulfuric acid react to form ammonium sulfate. Determine the starting mass of each reactant if 20.3 g of ammonium sulfate is produced and 5.89g of sulfuric acid remains unreacted.

Answers

Answer: The mass of ammonia is 5.236 g and that of sulfuric acid is 15.064 g

Explanation:

Calculating the mass of ammonia:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of ammonium sulfate = 20.3 g

Molar mass of ammonium sulfate = 132.14 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonium sulfate}=(20.3g)/(132.14g/mol)=0.154mol$

The chemical equation for the reaction of ammonia and sulfuric acid follows:

$2NH_3+H_2SO_4\rightarrow (NH_4)_2SO_4$

As, sulfuric acid remains unreacted, which means it is an excess reagent and its starting mass cannot be determined from ammonium sulfate.

By Stoichiometry of the reaction:

1 mole of ammonium sulfate is produced by 2 moles of ammonia.

So, 0.154 moles of ammonium sulfate is produced by = $(2)/(1)* 0.154=0.308mol$ of ammonia.

Now, calculating the mass of ammonia from equation 1, we get:

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 0.308 moles

Putting values in equation 1, we get:

$0.308mol=\frac{\text{Mass of ammonia}}{17g/mol}\n\n\text{Mass of ammonia}=5.236g$

Calculating the mass of sulfuric acid

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

Let the mass of sulfuric acid be 'x' grams

We are given:

Mass of ammonium sulfate = 20.3 grams

Mass of ammonia = 5.236 grams

Total mass on reactant side = 5.236 + x

Total mass on product side = 20.3 g

So, by applying law of conservation of mass, we get:

$5.236+x=20.3\n\nx=15.064g$

Hence, the mass of ammonia is 5.236 g and that of sulfuric acid is 15.064 g

2NH₃ + H₂SO₄ ⇒ (NH₄)₂SO₄
34.06g : 98.08g : 132.14g
x y 20.03g

x = [20.03g*34.06g]/132.14g = 5.16g

5.16g + y = 20.03g ⇔ from stoichiometric ratio
20.03g - 5.16g = y
y = 14,87g + 5.89g = 20.76g ⇔ 5.89g of sulfuric(VI) acid remains unreacted

Starting mass:

mNH₃ = 5.16g
mH₂SO₄ = 20.76g

Which substances dissolves in pure water and produces a solution that is a good conductor of electricity?1) CaCl2
2) C6H12O6
3) N2
4) O2

Answers

Answer:

$CaCl_2$

Explanation:

We have to keep in mind that a good conductor would be a substance that has the ability to produce ions (charged species). These species are responsible for the electricity conduction on water.

Therefore, we have to check which ones of these compounds will produce ions in water. For $N_2$ and $O_2$ we have diatomic molecules (also gases) this type of molecules don't have the ability to produce ions due to his simplicity, they already are very stable.

In the sugar molecule $C_6H_12O_6$ we have a covalent compound. The ions are produced by ionic compounds. So, this molecule can be discarded too.

For $CaCl_2$ we have an ionic compound that have the ability to produce ions:

$CaCl_2~~->~~Ca^+^2~+~2Cl^-$

So, this is the compound that would be a good conductor due to the $Ca^+^2$ and the $Cl^-$ ions.

CaCl2 is a good conductor of electricity when dissolved in water.

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