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raketka [301]
2 years ago
14

1:( Que estudia la química ?

Chemistry
1 answer:
AURORKA [14]2 years ago
4 0
Jfiaucuqixhbwixywhixyebw
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What does water undergo when it turns<br> to gas from liquid on surface of a<br> sidewalk?
Svetradugi [14.3K]

Answer:

Boiling - when the liquid is heated to a gas.

Evaporating - when the air temperature is hotter than the surface of the liquid so the water turns into water vapor or a gas.

Explanation:

6 0
3 years ago
Write the chemical formula for the cation present in the aqueous solution of cuso4.
Inessa05 [86]

The ionic compounds on dissolving in aqueous solution dissociate into their respective ions that are cations and anions.

The ions present in the aqueous solution of CuSO_4 are Cu^{2+} and SO_{4}^{2-}.

So, the chemical formula of the cation present in the aqueous solution of CuSO_4 is Cu^{2+}.

4 0
3 years ago
Match the humidities with their description.
Sloan [31]
1. C
2. D
3. A
4. B

Hope this helped! Please brainliest!
8 0
3 years ago
In the reaction of aluminum metal and oxygen to make aluminum oxide, how many grams of oxygen gas will react with 2.2 moles alum
sweet-ann [11.9K]

Answer:

52.8 g of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4Al + 3O2 —> 2Al2O3

From the balanced equation above,

4 moles of Al reacted with 3 moles of O2 to produce 2 moles of Al2O3

Next, we shall determine the number of mole of O2 needed to react with 2.2 moles of Al. This can be obtained as follow:

From the balanced equation above,

4 moles of Al reacted with 3 moles of O2.

Therefore, 2.2 moles of Al will react with = (2.2 × 3)/4 = 1.65 moles of O2.

Thus, 1.65 moles of O2 is needed for the reaction.

Finally, we shall determine the mass of O2 needed as shown below:

Mole of O2 = 1.65 moles

Molar mass of O2 = 2 × 16= 32 g/mol

Mass of O2 =?

Mole = mass/Molar mass

1.65 = mass of O2 /32

Cross multiply

Mass of O2 = 1.65 × 32

Mass of O2 = 52.8 g

Therefore, 52.8 g of O2 is needed for the reaction.

4 0
3 years ago
You are given 10ml (M) 20 Naoh solution in a conical flask and asked to titrate with (M) 20 Hcl and (M) 20 H2so4 separately. cal
Solnce55 [7]

Answer:

n_{HCl}=0.2molHCl\\n_{H_2SO_4}=0.1molH_2SO_4

Explanation:

Hello!

In this case, since the reactions between NaOH and the acids are:

NaOH+HCl\rightarrow NaCl+H_2O\\\\2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

Whereas we can see the 1:1 and 2:1 mole ratios between NaOH and HCl and H2SO4 respectively. In such a way, at the equivalence point we realize that:

n_{HCl}=n_{NaOH}=V_{NaOH}M_{NaOH}=0.01L*20mol/L=0.2molHCl\\\\2n_{H_2SO_4}=n_{NaOH}\\\\n_{H_2SO_4}=\frac{1}{2} V_{NaOH}M_{NaOH}=\frac{0.01L*20mol/L}{2} =0.1molH_2SO_4

Best regards!

8 0
2 years ago
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