How many milliliters are equal to 0.399195L

how many grams of calcium sulfate are produced from 10 grams of calcium nitrate and how many grams of calcium sulfate are produc

AlekseyPX

Answer: 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Ca(NO_3)_2=\frac{10g}{164g/mol}=0.061moles$

$\text{Moles of} Li_2SO_4=\frac{10g}{110g/mol}=0.091moles$

$Ca(NO_3)_2+Li_2SO_4\rightarrow 2LiNO_3+CaSO_4$

According to stoichiometry :

1 mole of $Ca(NO_3)_2$ require = 1 mole of $Li_2SO_4$

Thus 0.061 moles of $Ca(NO_3)_2$ will require= $\frac{1}{1}\times 0.061=0.061moles$ of $Li_2SO_4$

Thus $Ca(NO_3)_2$ is the limiting reagent as it limits the formation of product and $Li_2SO_4$ is the excess reagent.

As 1 mole of $Ca(NO_3)_2$ give = 1 mole of $CaSO_4$

Thus 0.061 moles of $Ca(NO_3)_2$ give = $\frac{1}{1}\times 0.061=0.061moles$ of $CaSO_4$

Mass of $CaSO_4=moles\times {\text {Molar mass}}=0.061moles\times 136g/mol=8.30g$

Thus 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.

6 0

3 years ago

What is a limiting factor?

alukav5142 [94]

A limiting factor helps an organism outcompete other organisms i think that’s the answer

5 0

2 years ago

PLEASE HELP FAST IM ABOUT TO FAIL PLEASE HELP PLEASE HELP FAST HELP HELP THIS IS DESPRATE

MariettaO [177]

1. 0.240 liters of water would be needed to dissolve 21.6 g of lithium nitrate to make a 1.3 M (molar) solution.

2. 2.9 M is the molarity of a solution made of 215.1 g of HCl is dissolved to make 2.0 L of solution.

3.83.3 ml of concentrated 18 M H2SO4 is needed to prepare 250.0 mL of a 6.0 M solution.

4. 135 ml of stock HBr will be required to dilute the solution.

5. 150 ml of water should be added to 50.0 mL of 12 M hydrochloric acid to make a 4.0 M solution

6. The pH of the resulting solution is 13.89

Explanation:

The formula used in solving the problems is

number of moles= $\frac{mass}{atomic mass of one mole}$ 1st equation

molarity = $\frac{number of moles}{volume}$ 2nd equation

Dilution formula

M1V1 = M2V2 3rd equation

1. Data given

mass of Lithium nitrate = 21.6 grams

atomic mass of on emole lithium nitrate = 68.946 gram/mole

Molarity is given as 1.3 M

VOLUME=?

Calculate the number of moles using equation 1

n = $\frac{21.6}{68.946}$

= 0.313 moles of lithium nitrate.

volume is calculated by applying equation 2.

volume = $\frac{0.313}{1.3}$

= 0.240 litres of water will be used.

2. Data given:

mass of HCl = 215.1 gram

atomic mass of HCl = 36.46 gram/mole

volume = 2 litres

molarity = ?

using equation 1 number of moles calculated

number of moles = $\frac{215.1}{36.46}$

number of moles of HCl = 5.899 moles

molarity is calculated by using equation 2

M = $\frac{5.899}{2}$

= 2.9 M is the molarity of the solution of 2 litre HCl.

3. data given:

molarity of H2SO4 = 18 M

Solution to be made 250 ml of 6 M

USING EQUATION 3

18 x V1= 250 x 6

V1 = 83.3 ml of concentrated 18 M H2SO4 will be required.

4. data given:

M1= 10M, V1 =?, M2= 3 ,V2= 450 ml

applying the equation 3

10 x VI = 3x 450

V1 = 135 ml of stock HBr will be required.

5. Data given:

V1 = 50 ml

M1= 12 M

V2=?

M2= 4

applying the equation 3

50 x 12 = 4 x v2

V2 = 150 ml.

6. data given:

HCl + NaOH ⇒ NaCl + H20

molarity of NaOH = 0.525 M

volume of NaOH = 25 ml

molarity of acid HCl= 75 ml

volume of HCl = 0.335 ml

pH=?

Number of moles of NaOH and HCl is calculated by using equation 1 and converting volume in litres

moles of NaOH = 0.0131

moles of HCl= 0.025 moles

The ratio of moles is 1:1 . To find the unreacted moles of acid and base which does not participated in neutralization so the difference of number of moles of acid minus number of moles of base is taken.

difference of moles = 0.0119 moles ( NaOH moles is more)

Molarity can be calculated by using equation 1 in (25 +75 ml) litre of solution

molarity = $\frac{0.0119}{0.1}$

= 0.11 M (pOH Concentration)

14 = pH + pOH

pH = 14 - 0.11

pH = 13.89

3 0

3 years ago

If 20.0 ml of glacial acetic acid (pure hc2h3o2) is diluted to 1.40 l with water, what is the ph of the resulting solution? the

Y_Kistochka [10]

By using the formula, mass = density x volume, we calculate mass in grams

20.0 mL CH₃COOH x (1.05 g / mL) = 21.0 g CH₃COOH

To find the moles, molar mass of CH₃COOH = 60.05g/mol

21.0 g CH₃COOH x (1 mole CH₃COOH / 60.05 g CH₃COOH) = 0.350 moles CH₃COOH

To find molarity,

[CH₃COOH] = moles CH₃COOH / L of solution = 0.350 / 1.40 = 0.250 M CH₃COOH

When CH₃COOH is dissolved in water, it produces small and equal amounts of H₃O⁺+ and C₂H₃O₂⁻.

Molarity , CH₃COOH + H₂O <==> H₃O⁺ + C₂H₃O₂⁻

Initial 0.250 0 0 
Change -x x x
Equilibrium 0.250-x x x

Kₐ = [H₃O⁺][C₂H₃O₂⁻] / [HC₂H₃O₂] = (x)(x) / (0.250-x) = 1.8 x 10⁻⁵

Since Kₐ is relatively small, we can neglect the -x term after 0.250 to simplify

x² / 0.250 = 1.8 x 10⁻⁵

x² = 4.5 x 10⁻⁶

x = 2.1 x 10⁻³ = [H₃O⁺]

pH = -log [H₃O⁺] = -log (2.1 x 10⁻³) = 2.68

8 0

3 years ago

In a stadium, fans stand up and sit down to produce a wave across the stadium. This type of wave where the material travels perp

elena-14-01-66 [18.8K]

Transverse wave is a wave vibrating at right angles to the direction of its propagation.

In short, Your Answer would be Option B

Hope this helps!

7 0

3 years ago

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