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aksik [14]
4 years ago
13

Explain the steps involved In the purification and treatment of drinking water​

Chemistry
2 answers:
ch4aika [34]4 years ago
7 0

Answer:

<h3>.Get a clean container or kettle.</h3><h3>Pour in clean water</h3><h3>And then start boiling.</h3>

<h3>Or u can put water guard in the cooked H2O</h3>
Natali [406]4 years ago
6 0

Answer:

1) Filter to remove insoluble substances from the water

2) Use carbon to remove odor.

3) Chlorine is used to kill bacteria.

Explanation:

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Glucose is the monomer in the natural polymers ____________________ and cellulose.
Olin [163]

Answer:

Starch

I hope this helps

5 0
3 years ago
SI unit for measuring distance is
MrMuchimi
The SI unit for measuring distance is the meter.
I attached a table of SI measurements for you :)
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6 0
3 years ago
20 POINTS!!!
nasty-shy [4]

Answer:

300g of water because its a larger amount

Explanation:

3 0
3 years ago
What is the valence electron of manganese?​
DIA [1.3K]

Answer:

25th electron

Explanation:

the last electron is the valence electron. Assuming it has equal numbers of protons and electrons, then the 25th electron is the valence.

6 0
3 years ago
Read 2 more answers
Calculate the standard molar enthalpy of formation, in kJ/mol, of NO(g) from the following data:
Paraphin [41]

The above question is incomplete, here is the complete question:

Calculate the standard molar enthalpy of formation of NO(g) from the following data at 298 K:

N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o = 66.4 kJ

2NO(g) + O_2\rightarrow 2NO_2(g),\Delta H^o= -114.1 kJ

Answer:

The standard molar enthalpy of formation of NO is 90.25 kJ/mol.

Explanation:

N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o_{1} = 66.4 kJ

2NO(g) + O_2\rightarrow 2NO_2(g),\Delta H^o_{2} = -114.1 kJ

To calculate the standard molar enthalpy of formation

N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?...[3]

Using Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

[1] - [2] = [3]

N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?

\Delta H^o_{3} =\Delta H^o_{1} - \Delta H^o_{2}

\Delta H^o_{3}=66.4 kJ - [ -114.1 kJ] = 180.5 kJ

According to reaction [3], 1 mole of nitrogen gas and 1 mole of oxygen gas gives 2 mole of nitrogen monoxide, So, the standard molar enthalpy of formation of 1 mole of NO gas :

=\frac{\Delta H^o_{3}}{2 mol}

=\frac{180.5 kJ}{2 mol}=90.25 kJ/mol

3 0
3 years ago
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