Ask question

Ask question

All categories

Murljashka [212]

2 years ago

7

Observations:Sodium hydroxide reacts with coper sulfate

1 answer:

vaieri [72.5K]2 years ago

6 0

Answer:

copper sulphate reacts with sodium hrydroxide to form a blue preipitate of copper hrydroxide and soduim hrydroxide.

You might be interested in

Which substance is the oxidizing agent in this reaction? 2CuO+C→2Cu+CO2 Express your answer as a chemical formula.

Alja [10]

The oxidizing agent is the one that is reduced in the reaction. In this reaction, the charge of Cu falls from +2 to zero charge (neutral atom in the right side). Hence, CuO is the oxidizing agent. The reducing agent, the one being oxidized is carbon from zero charge to +4. The answer is CuO.

3 0

3 years ago

A student dissolves 4 g of salt into 10 g of water. The total weight of the solution is 14 g. The student then heats the solutio

shtirl [24]

C. or B. ok :) !! ! Hopes this hepls

3 0

3 years ago

Why are heating vents usually placed on the floor of a home?

Leno4ka [110]

Because in heat rises faster on the floor than on the ceiling or side of a wall.

7 0

3 years ago

Read 2 more answers

What type of bonding holds the atoms in an h2o molecule together?

Ivanshal [37]

Covalent bond two non metals

3 0

3 years ago

If a pork roast must absorb kJ to fully cook, and if only 10% of the heat produced by the barbecue is actually absorbed by the r

kolezko [41]

Answer:

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

Explanation:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g),\Delta H_{rxn}=-2044 kJ/mol (assuming)$

Energy absorbed by pork,E = $1.6\times 10^3 kJ$ (assuming)

Total energy produced by barbecue = Q

Percentage of energy absorbed by pork = 10%

$10\%=\frac{E}{Q}\times 100$

$Q=\frac{E}{10}\times 100=\frac{1.6\times 10^3 kJ}{10}\times 100=1.6\times 10^4 kJ$

Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.

Q = $-1.6\times 10^4 kJ$

Moles of propane burnt to produce Q energy =n

$n\times \Delta H_{rxn}=Q$

$n=\frac{Q}{\Delta H_{rxn}}=\frac{-1.6\times 10^4 kJ}{-2044 kJ/mol}=7.83 mol$

According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:

$\frac{1}{3}\times 7.83 mol=23.49 mol$ carbon dioxide gas.

Mass of 23.49 moles of carbon dioxide gas:

23.49 mol × 44 g/mol =1,033.56 g

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

8 0

3 years ago