All categories

2 years ago

CuSO4 . 5H2O is dehydrated to form Copper (ii) sulfate. Is this a physical or chemical change? Explain.

Chemistry

1 answer:

Paraphin [41]2 years ago

3 0

Answer:

It is a physical change.

Explanation:

• This is because when anhydrated copper (ii) sulfate is heated, the water evaporates to form anhydrous copper (ii) sulfate.

• evaporation is a physical change

$.$

You might be interested in

Which of the following are examples of conservable quantities? a. potential energy and length c. mechanical energy and mass b. m

Blababa [14]

Answer:

Explanation:

7 0

2 years ago

Write the net ionic equation for the reaction between hydrocyanic acid and potassium hydroxide. Do not include states such as (a

eimsori [14]

Answer:

The net ionic equation is as follows:

HCN(aq) + OH-(aq) ----> H20(l) + CN-(aq)

Explanation:

The reaction between Hydrocyanic acid, HCN, and sodium hydroxide is a neutralization reaction between a weak acid and a strong base.

Hydrocyanic acid being a weak acid ionizes only slightly, while sodium hydroxide being a strong base ionizes completely. The equation for the reaction is given below:

A. HCN(aq) + NaOH-(aq) ----> NaCN(aq) + H2O(l)

Since Hydrocyanic acid is written in the aqueous form as it ionizes only slightly and the ionic equation is given below:

HCN(aq) + Na+(aq)+OH-(aq) ----> Na+(aq)+CN-(aq) + H2O(l)

Na+ being a spectator ion is removed from the net ionic equation given below:

HCN(aq) + OH-(aq) ----> H20(l) + CN-(aq)

4 0

2 years ago

If the density of an object is 5.2g/cm3, and it’s volume is 3.7 cm3, what is it’s mass?

lord [1]

Here's the equation you use: Density = mass/volume

1) 5.2g/cm^3 = m/3.7cm^3

2) m = 5.2g/cm^3 x 3.7cm^3

3) m = 19.24g

You can check the answer by plugging it in

19.24g/3.7cm^3
= 5.2g/cm^3

5 0

2 years ago

Limiting Reactants—————-

denis-greek [22]

Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

Mg: 24.301;
O: 15.999.

How many moles of MgO will be produced if Mg is the limiting reactant?

Number of moles of Mg:

$\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}$ .

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

$\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}$ .

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. $2\times 0.270642 = 0.541284\;\text{mol}$ of MgO will be produced if O₂ is in excess.