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yulyashka [42]
3 years ago
7

Besides water, what is the product of a Neutalization Reaction between HNO3 and Al(OH)3?

Chemistry
2 answers:
aksik [14]3 years ago
7 0
Answer would be c. AINO3
Alexxandr [17]3 years ago
5 0

Explanation:

Besides water what is the product of a Neutralization Reaction between HNO3 and Al(OH)3

AINO3

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Two protons and two neutrons are released as a result of this reaction. mc017-1.jpgRn mc017-2.jpgPo + ? Which particle is releas
Dovator [93]
The nuclear reaction occurring is known as alpha-decay, and during this process, an alpha particle is released from a heavy radioactive nucleus to form a lighter more stable nucleus. The alpha particle is equivalent to a helium nucleus, which means it contains 2 protons and two neutrons (net charge of +2)
The decay equation is:
Rn → Po + α
5 0
4 years ago
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A gas that effuses through a porous cylinder 1.87 times faster than chlorine gas. what is the molar mass and identity.
Rufina [12.5K]
From the Graham's law of effusion;
R1/R2 = √MM2/√MM1
Molar mass of chlorine gas is 71
Therefore;
1.87= √ 71 /√mm1
= 1.87² = 71/mm1 
mm1 = 71/1.87²
         = 71/3.4969
         = 20.3
Thus, the molar mass of the other gas is 20.3 , and i think the gas is neon
6 0
4 years ago
A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction t
andreev551 [17]

Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)

So, moles of product formed are calculated as follows.

\frac{3}{2} \times 0.17 mol \\= 0.255 mol

Hence, the given data is as follows.

n_{1} = 0.17 mol,      n_{2} = 0.255 mol

V_{1} = 19.5 L,         V_{2} = ?

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}

Substitute the values into above formula as follows.

\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L

Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.

8 0
3 years ago
How many Oxygen atoms in: 5H2SO3
astra-53 [7]

Answer:

20

Explanation: I think

5 0
3 years ago
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What is the mass of 1.45 moles of silver sulfate?
tatuchka [14]

Answer:

449.5 g

Explanation:

Silver sulfate- Ag2SO4

M(Ag)=107 g/mol => M(Ag2)=214 g/mol

M(S)=32 g/mol

M(O)=16 g/mol => M(O4)=64 g/mol

M(Ag2SO4)=310 g/mol

n=1.45 mol

m(Ag2SO4)=M(Ag2SO4)*n=310 g/mol *1.45 mol= 449.5 g

3 0
3 years ago
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