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xeze [42]
3 years ago
9

What is the formula of copper sulphate?​

Chemistry
2 answers:
Diano4ka-milaya [45]3 years ago
8 0

_____________________________

Answer: The formula of Copper Sulphate is CuSO4. It contains one atom of Copper, one atom of Sulphur and four atom of Oxygen. It is an acidic salt which is formed by the neutralization reaction of Copper Hydroxide and Sulphuric acid. It turns blue litmus paper into red which proves that it is acidic in nature. It is colourless compound but when it contains some water of crystallization then it will turns blue in colour.

_____________________________

mylen [45]3 years ago
4 0

COPPER SULPHATE =CuSO4

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All plants and animals need energy in order to survive. The diagram above shows how energy gets transferred among living things
Andre45 [30]
I believe it would be A, frogs and spiders, because frogs and spiders both eat grasshoppers. But I cannot see the diagram it talks about, so it's harder to tell.
8 0
2 years ago
. Calculate the mass of O2 produced if 3.450 g potassium chlorate is completely decomposed by heating in presence of a catalyst
Vesnalui [34]
 <span>2 KClO3(s) → 3 O2(g) + 2 KCl(s) 

</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates. 
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 =  3.45 / 122.55 = 0.028

Moles of O2 produce = \frac{3}{2} \times 0.028

= 0.042 moles

molar mass of O2 = 32

so, mass of O2 = 32 x 0.042  = 1.35 g



5 0
2 years ago
Write balanced chemical equations for the following reactions.
VLD [36.1K]

❃ The following points should be kept in mind to write and balance a chemical equation :

Step 1 : Write the molecular formula of all the reactants and products correctly.

Step 2 : Separate reactants and products by a sign of arrow. If reactants or products are more than one, connect them by a sign of a plus.

Step 3 : Balance the atoms of O and H at last [ The atoms used at many places in an equation should be balanced at last ]. For balancing , the number should be added as coefficient i.e in the front of the molecules.

[ Remember those substance that take part in a chemical reaction are called reactants. Likewise , those substances which are formed after a chemical reaction are called products ]

\large{ \tt{❁ \: LET'S \: GET \: STARTED}} :

1. Carbon disulfide + Oxygen gas gives carbon dioxide + Sulfur dioxide.

Step 1 : The molecular formula of carbon disulfide is CS₂ , molecular formula of Oxygen gas is 0₂ [ Since oxygen is a diatomic element ] molecular formula of carbon dioxide is CO₂ and molecular formula of sulfur dioxide is SO₂.

Step 2 : CS₂ + O₂ ⟶ CO₂ + SO₂

Step 3 : In the reactant side , there is two ' S ' but on the other side , there is one ' S '. So , add 2 as a coefficient before S on the product side. Now , There are two ' O ' in the reactant side but six ' O ' in the product side. So , add 3 as a coefficient before O on the reactant side. Now , there are equal atom of C , S and O on both sides

i.e CS₂ + 3O₂ ⟶ CO₂ + 2SO₂

Answer : \boxed{ \tt{CS₂ + 3O₂ ⟶ CO₂ + 2SO₂}}

-----------------------------------------

2. Silver + nitric acid gives silver nitrate + nitrogen dioxide + water

Step 1 : The molecular formula of Silver is Ag, molecular formula of nitric acid is HNO₃ , molecular formula of Silver nitrate is Ag ( No₃ ) , molecular formula of nitrogen dioxide is NO₂ and molecular formula of water is H₂O.

Step 2 : Ag + HNO₃ ⟶ Ag ( NO₃ ) + NO₂ + H₂O

Step 3 : In the reactant side , There is one ' H ' but on the other side , there are two ' H '. Now add 2 before H on the reactant side. There are equal atom of ' Ag ' , ' H ' , ' N ' , and ' O '.

i.e Ag + 2HNO₃ ⟶ Ag ( NO₃ ) + NO₂ + H₂O

Answer : \boxed{ \tt{Ag + 2HNO₃ ⟶ Ag ( NO₃ ) + NO₂ + H₂O }}

  • Yay! We're done ! :)

- The last step is a bit more confusing I guess. So , which balancing , count the atoms in following ways :

  • The number written at the right lower corner of an atom is counted for that atom only. For example : In MgSO₄ , there are one ' Mg ' , one ' S ' and four ' O '

  • The number written at the right lower corner of a bracket is for all the atoms enclosed in the bracket. For example : In Al₂ ( SiO₃ ) has two Al , three ' S ' and nine ' O '.

  • The coefficient number is for all the atoms of the molecule. For example , in 2 Al ₂( SiO₃ )₃ , there are four ' Al ' , six ' Al ' and eighteen ' O '.

- Hope this helps! :)

4 0
2 years ago
Read 2 more answers
What does it mean when your teacher wants you to round 143 to two significant figures
zhenek [66]

Answer:The answer is 3...I think

Explanation:

143

Sig Figs

3

Decimals

0

Scientific Notation

1.43 × 102

5 0
3 years ago
A sample of quartz is put into a calorimeter (see sketch at right) that contains of water. The quartz sample starts off at and t
pashok25 [27]

Answer:

0.71 J/g°C

Explanation:

Here is the complete question

thermometer A 51.9 g sample of quartz is put into a calorimeter (see sketch at right) that contains 300.0 g of water. The quartz sample starts off at 97.8 °C and the temperature of the water starts off at 17.0 °C. When the temperature of the water stops changing it's 19.3 °C. The pressure remains constant at 1 atm. insulated container water sample Calculate the specific heat capacity of quartz according to this experiment. Be sure your answer is rounded to 2 significant digits. a calorimeter g °C

Solution

Since the temperature of the water increases from 17.0 °C to 19.3 °C, it means that it loses heat. Also, the final temperature of the quartz equals the final temperature of the water 19.3 °C. Since the quartz temperature decreases from 97.8 °C to 19.3 °C it loses heat.

So, heat lost by quartz, Q = heat gained by water, Q'

-Q = Q'

-mc(θ₂ - θ₁) = m'c'(θ₂ - θ₃) where m = mass of quartz = 51.9 g, c = specific heat capacity of quartz, θ₁ = initial temperature of quartz = 97.8 °C, θ₂ = final temperature of quartz = 19.3 °C, m' = mass of water = 300 g, c = specific heat capacity of water = 4.2 J/g °C , θ₃ = initial temperature of water = 17.0 °C, θ₂ = final temperature of water = 19.3 °C

Making c subject of the formula, we have

c = -m'c'(θ₂ - θ₃)/m(θ₂ - θ₁)

Substituting the values of the variables into the equation, we have

c = -300 g × 4.2 J/g °C(19.3 °C - 17.0 °C)/51.9 g(19.3 °C - 97.8 °C)

c = -1260 J/°C(2.3 °C)/51.9 g(-78.5 °C)

c = -2898 J/-4074.15 g°C

c = 0.711 J/g°C

c ≅ 0.71 J/g°C to 2 significant digits

5 0
3 years ago
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