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guajiro [1.7K]
3 years ago
9

Is poh level a acid or base

Chemistry
1 answer:
Zolol [24]3 years ago
8 0
It’s a base because it has hydroxide (oh) and not just hydrogen.
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29. Which is the electron configuration of an atom
Feliz [49]
2 is right am 70 percent sure
8 0
3 years ago
Jimmy mixes 2 chemicals in the lab. The chemicals change color and they get hot. The rise in temperature
gladu [14]

Answer:

A chemical reaction.

Explanation:

A change in temperature is evidence of a chemical reaction.

Also: They are chemicals...

3 0
3 years ago
In the second step of this reaction, isocyanic acid reacts to form melamine and carbon dioxide: HNCO(l)→C3N3(NH2)3(l)+CO2(g) Bal
kykrilka [37]

Answer:

The coefficients are 6, 1, 3

Explanation:

HNCO →C3N3(NH2)3 + CO2

From the above equation, there are a total of 6 atoms of nitrogen on the right side and 1atom on the left. It can be balance by putting 6 in front of HNCO as shown below:

6HNCO → C3N3(NH2)3 + CO2

Now there are 6 atoms of carbon on the left side and 4 atoms on the right side. It can be balance by putting 3 in front of CO2 as shown below:

6HNCO → C3N3(NH2)3 + 3CO2

Now the equation is balanced as the numbers of atoms of the different elements on both sides of the equation are the same.

The coefficients are 6, 1, 3

8 0
3 years ago
Which expression is equal to the number of grams in 2.43 kilograms
KiRa [710]
1 kg = 1000g
2.43 kg *1000g/1kg = 2430 g
3 0
4 years ago
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

4 0
3 years ago
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