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1 year ago

How many moles of tungsten (W,183.85 g/lol are in 415 grams of tungsten?

Chemistry

1 answer:

vladimir1956 [14]1 year ago

7 0

Given mass of tungsten, W = 415 g

Molar mass of tungsten, W = 183.85 g/mol

Calculating moles of tungsten from mass and molar mass:

$415 g * \frac{1 mol}{183.85 g} = 2.26 mol W$

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(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW

Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M InF₃. </em>Thus:

[In⁺³] = 2.3x10⁻³M InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M In⁺³

[F⁻] = 2.3x10⁻³M InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, <em>solubility is 6.31x10⁻⁶M</em>

3 0

2 years ago

Calculate the molar mass of a 2.89 g gas at 346 ml, a temperature of 28.3 degrees Celsius, and a pressure of 760 mmHg.

malfutka [58]

The molar mass of gas = 206.36 g/mol

<h3>Further explanation</h3>

In general, the gas equation can be written

$\large{\boxed{\bold{PV=nRT}}}$

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.082 l.atm / mol K

T = temperature, Kelvin

mass (m)= 2.89 g

volume(V) = 346 ml = 0.346 L

T = 28.3 C + 273 = 301.3 K

P = 760 mmHg=1 atm

The molar mass (M) :

$\tt PV=\dfrac{m}{M}RT\\\\M=\dfrac{mRT}{PV}\\\\M=\dfrac{2.89\times 0.082\times 301.3}{1\times 0.346}\\\\M=206.36~g/mol$

8 0

1 year ago

Gina wants to use models to better understand how the types of bonds in a molecule relate to the presence of geometric isomers.

vazorg [7]

The answer to this question is D! The ball and stick model! Hope this helps :)

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1 year ago

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A pharmacist–herbalist mixed 100 g lots o St. John’s wort containing the ollowing percentages o the active component hypericin:

Anna [14]

Answer:

strength of hypericin in mixture = 0.42 %

Explanation:

given data

each lot = 100 g

active component hypericin = 0.3%, 0.7%, and 0.25%

solution

we get here percent strength o hypericin in the mixture that is

Hypericin contribution lot 1 = $\frac{0.3}{100}$ × 100

Hypericin contribution lot 1 = 0.3 g

and

Hypericin contribution lot 2 = $\frac{0.3}{100}$ × 100

Hypericin contribution lot 2 = 0.7 g

and

Hypericin contribution lot 3 = $\frac{0.25}{100}$ × 100

Hypericin contribution lot 3 = 0.25 g

total 300 g mixture of hypericin contain = 0.3 g + 0.7 g + 0.25 g

total 300 g mixture of hypericin = 1.25 g

so here percent strength o hypericin in mixture is

strength of hypericin in mixture = $\frac{1.25}{300}$ × 100

strength of hypericin in mixture = 0.42 %

5 0

1 year ago

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r

Scilla [17]

I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.

After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.

6 0

2 years ago