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2 years ago

PLS HELP I GIVE BRANLIEST !!!

Chemistry

1 answer:

yarga [219]2 years ago

6 0

Explanation:

D. The total mass of the reactants equals the total mass of the

products.

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Calculate the number of milliliters of 0.710 M Ba(OH)2 required to precipitate all of the Mn2+ ions in 161 mL of 0.796 M KMnO4 s

USPshnik [31]

Answer: The volume of barium hydroxide is 183 mL.

Explanation:

To calculate the moles of cadmium nitrate, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of $MnSO_4$ = 0.796 M

Volume of $MnSO_4$ = 161 mL = 0.161 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.796mol/L=\frac{\text{Moles of }MnSO_4}{0.161L}\\\\\text{Moles of }MnSO_4=0.13mol$

The chemical equation for the reaction of manganese sulfate and barium hydroxide follows:

$MnSO_4(aq.)+Ba(OH)_2(aq.)\rightarrow Mn(OH)_2(s)+BaSO_4(aq.)$

By Stoichiometry of the reaction:

1 mole of manganese sulfate reacts with 1 mole of barium hydroxide.

So, 0.13 moles of manganese sulfate will react with = $\frac{1}{1}\times 0.13=0.13mol$ of barium hydroxide

Now, calculating the volume of barium hydroxide by using equation 1, we get:

Moles of barium hydroxide = 0.13 moles

Molarity of barium hydroxide = 0.710 M

Putting values in equation 1, we get:

$0.710mol/L=\frac{0.13mol}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=0.183L$

Converting this into milliliters, we use the conversion factor:

1 L = 1000 mL

So, $0.183L=0.183\times 1000=183mL$

Hence, the volume of barium hydroxide is 183 mL.

7 0

3 years ago

How many moles are in 2.5L of 1.75 M Na2CO3

mixas84 [53]

Answer: There are 4.375 moles in 2.5 L of 1.75 M $Na_2CO_3$

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of solution = 1.75 M

Volume of solution = 2.5 L

Putting values in equation , we get:

$1.75M=\frac{\text{Moles of} Na_2CO_3}{2.5L}\\\\\text{Moles of }Na_2CO_3=1.75mol/L\times 2.5L=4.375mol$

6 0

2 years ago

How many grams of magnesium metal will react completely with 5.2 liters of 4.0 M HCl?

neonofarm [45]

1) Find the number of mols of HCl in 5.2 liters of 4.0M solution:

n = M*V(L) = 4.0 mol/L * 5.2 L = 20.8 mol

2) Find the number of mols of Mg that will react with 20.8 mol of HCl, using the coefficients of the balanced equation

[1mol Mg / 2 mol HCl] * 20.8 mol HCl = 10.4 mol Mg

3) Transform mol to mass using the atomic mass:

10.4 mol Mg * 24.3 g/mol = 252.7 g of Mg.

4 0

2 years ago

Which will dissolve at the fastest rate?

Nezavi [6.7K]

I need help on this one too I think it might be (b)

7 0

2 years ago

Hydrofluoric acid, hf, has a ka of 6.8 × 10−4. what are [h3o+], [f−], and [oh−] in 0.710 m hf?

STALIN [3.7K]

Answer:

[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.

Explanation:

For a weak acid like HF, the dissociation of HF will be:

HF + H₂O ⇄ H₃O⁺ + F⁻.

[H₃O⁺] = [F⁻].

∵ [H₃O⁺] = √Ka.C,

Ka = 6.8 x 10⁻⁴, C = 0.710 M.

∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.

∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(2.2 x 10⁻²) = 4.55 x 10⁻¹³.

6 0

2 years ago