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3 years ago

How many molecules are found in 90 grams of Al(OH)3?

Chemistry

1 answer:

jarptica [38.1K]3 years ago

6 0

Answer:

6.95 x 10²³ molecules/particles

Explanation:

First we need to find the total Empirical Mass. We can find this by adding each element's mass together.

Al = 26.982,

O = 15.999

H = 1.008

26.982 + 3(15.999) + 3(1.008) = 78.003.

Now we divide by the mass given (90 grams).

90/78.003 = 1.153801777.

We then take that number and multiply it by avogadro's number (6.022 x 10²³)

1.153801777 x avogadro's number = 6.95 x 10²³

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A. Chloroplasts

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C. Vacuoles

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3 years ago

Design an invitation you could use to calculate the density of a penny

Nataliya [291]

Firstly, the density of any substance is represented by the mass (amount of matter) as divided by the volume(amount of space). According to external websites, the mass of a penny is 2.5 grams.However, the volume of a penny is .35cm to the power of 3 (due to the thickness of the penny being extremely minimal.Thus the amount of density is extremely little). Therefore, the density of a penny is 0.875 g/cm cubed (dimensional analysis).As for an invention that could be used, that is possible with the usage of a series of measurements that can both calculate mass and volume and directly allocate that to attain density

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4 years ago

When a lithium ion and nitrate ion combine, the result is:

maxonik [38]

LiNO3; nitrate has a charge of -1, and lithium has a charge of +1

7 0

3 years ago

Read 2 more answers

A chemist fills a reaction vessel with 0.750 M lead (II) (Pb2+) aqueous solution, 0.232 M bromide (Br) aqueous solution, and 0.9

Ronch [10]

Answer:

The free energy = -20.46 KJ

Explanation:

given Data:

Pb²⁺ = 0.750 M

Br⁻ = 0.232 M

R = 8.314 Jk⁻¹mol⁻¹

T = 298K

The Gibb's free energy is calculated using the formula;

ΔG = ΔG° + RTlnQ -------------------------1

Where;

ΔG° = standard Gibb's freeenergy

R = Gas constant

Q = reaction quotient

T = temperature

The chemical reaction is given as;

Pb²⁺(aq) + 2Br⁻(aq) ⇄PbBr₂(s)

The ΔG°f are given as:

ΔG°f (PbBr₂) = -260.75 kj.mol⁻¹

ΔG°f (Pb²⁺) = -24.4 kj.mol⁻¹

ΔG°f (2Br⁻) = -103.97 kj.mol⁻¹

Calculating the standard gibb's free energy using the formula;

ΔG° = ξnpΔG°(product) - ξnrΔG°(reactant)

Substituting, we have;

ΔG° =[1mol*ΔG°f (PbBr₂)] - [1 mol *ΔG°f (Pb²⁺) +2mol *ΔG°f (2Br⁻)]

ΔG° =(1 *-260.75 kj.mol⁻¹) - (1* -24.4 kj.mol⁻¹) +(2*-103.97 kj.mol⁻¹)

= -260.75 + 232.34

= -28.41 kj

Calculating the reaction quotient Q using the formula;

Q = 1/[Pb²⁺ *(Br⁻)²]

= 1/(0.750 * 0.232²)

= 24.77

Substituting all the calculated values into equation 1, we have

ΔG = ΔG° + RTlnQ

ΔG = -28.41 + (8.414*10⁻³ * 298 * In 24.77)

= -28.41 +7.95

= -20. 46 kJ

Therefore, the free energy of reaction = -20.46 kJ

8 0

3 years ago

Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . If of water i

Kaylis [27]

Answer:

87.9%

Explanation:

Balanced Chemical Equation:

HCl + NaOH = NaCl + H2O

We are Given:

Mass of H2O = 9.17 g

Mass of HCl = 21.1 g

Mass of NaOH = 43.6 g

First, calculate the moles of both HCl and NaOH:

Moles of HCl: 21.1 g of HCl x 1 mole of HCl/36.46 g of HCl = 0.579 moles

Moles of NaOH: 43.6 g of NaOH x 1 mole of NaOH/40.00 g of NaOH = 1.09 moles

Here you calculate the mole of H2O from the moles of both HCl and NaOH using the balanced chemical equation:

Moles of H2O from the moles of HCl: 0.579 moles of HCl x 1 mole of H2O/1 mole of HCl = 0.579 moles

Moles of H2O from the moles of NaOH: 1.09 moles of HCl x 1 mole of H2O/1 mole of NaOH = 1.09 moles

From the calculations above, we can see that the limiting reagent is HCl because it produced the lower amount of moles of H2O. Therefore, we use 0.579 moles and NOT 1.09 moles to calculate the mass of H2O:

Mass of H2O: 0.579 moles of H2O x 18.02 g of H2O/1 mole of H2O = 10.43 g

% yield of H2O = actual yield/theoretical yield x 100= 9.17 g/10.43 g x 100 = 87.9%

3 0

4 years ago