Which of these is a unique property of acids?

A mole of lead will be heavier than a mole of oxygen because

tatuchka [14]

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8 0

3 years ago

Why is heat needed to get a reaction to occur

Ahat [919]

Answer:As temperature increases, reactions take place. Generally, higher temperatures mean faster reaction rates; as molecules move about more quickly, reactant molecules are more likely to interact, forming products.

Explanation:

4 0

3 years ago

Read 2 more answers

450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS

ExtremeBDS [4]

Answer:

1597.959 g

Explanation:

Given Data:

Amount of Cr₂(SO₄)₃ = 450 g

Amount of potassium phosphate K₃PO₄ = in Excess

grams of potassium sulfate K₂SO₄= ?

Solution

The Reaction will be

Cr₂(SO₄)₃ + 2K₃PO₄ ----------> 3K₂SO₄ + 2CrPO₄

Information that we have from reaction

Cr₂(SO₄)₃ + 2K₃PO₄ ----------> 3K₂SO₄ + 2CrPO₄

1 mol 2 mol 3 mol

we come to know from the above reaction that

1 mole of chromium(iii) sulfate (Cr₂(SO₄)₃) react with 2 mole of potassium phosphate (K₃PO₄) to produce 3 mole of K₂SO₄

We also know that

molar mass of Cr₂(SO₄)₃ = 147 g/mol

molar mass of K₃PO₄ = 212 g/mol

molar mass of K₂SO₄ = 174 g/mol

if we represent mole in grams then

Cr₂(SO₄)₃ + 2K₃PO₄ ----------> 3K₂SO₄ + 2CrPO₄

1 mol (147 g/mol) 2 mol (212 g/mol) 3 mol (174g/mol)

So, Now we have the following details

Cr₂(SO₄)₃ + 2K₃PO₄ ----------> 3K₂SO₄ + 2CrPO₄

147 g 424 g 522 g

So,

we come to know that 147 g of Cr₂(SO₄)₃ combine with 424 g of 2K₃PO₄ produce 522 g of K₂SO₄

So now we calculate that how many grams of potassium sulfate will be produced

Apply unity formula

147 g of Cr₂(SO₄)₃ ≅ 522 g of K₂SO₄

450 g of Cr₂(SO₄)₃ ≅ ? g of K₂SO₄

by doing cross multiplication

g of K₂SO₄ =522 g x 450 g / 147 g

g of K₂SO₄ = 1597.959 g

So the write answer is 1597.959 g

***Note: By calculation it is obvious that the correct answer is 1597.959 g

8 0

3 years ago

Which conclusion is supported by the information in the graph?

Molodets [167]

Answer:

A. The cost of producing a kilowatt of power with fuel cell will be less than $30 in 2015

4 0

2 years ago

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]

<u>Answer:</u> The volume of barium hydroxide is 183 mL.

<u>Explanation:</u>

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