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ruslelena [56]
3 years ago
10

Which compound could be added to hydroiodic acid to neutralize its acidic properties ?

Chemistry
2 answers:
netineya [11]3 years ago
6 0

methylamine, a weak base.

Juli2301 [7.4K]3 years ago
5 0

A base could be added so anything with a hydroxide or oxide group

E.g NaOH

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The metal that is used to hold stain glass pieces together is called __________.
inna [77]
The answer is D because most stained glass is held together with lead. The process of this is called leading.
4 0
2 years ago
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Balancee por tanteo las siguientes ecuaciones químicas. Escriba el nombre a reactantes y productos. H2O5 + H2O ---> HNO3 Na2O
Elan Coil [88]

Answer:

a. N₂O₅ + H₂O ⇒ 2 HNO₃ (pentóxido de dinitrógeno + agua ⇒ ácido nítrico)

b. Na₂O + H₂O ⇒ 2 NaOH (óxido de sodio + agua ⇒ hidróxido de sodio)

Explanation:

Tenemos que balancear, por el método de tanteo, las siguientes ecuaciones químicas.

a. En la primera reacción, el pentóxido de dinitrógeno reacciona con agua para formar ácido nítrico. Es una reacción de síntesis o combinación.

N₂O₅ + H₂O ⇒ HNO₃

Podremos obtener la ecuación balanceada si multiplicamos HNO₃ por 2.

N₂O₅ + H₂O ⇒ 2 HNO₃

b. En la segunda reacción, óxido de sodio reacciona con agua para formar hidróxido de sodio. Es una reacción de síntesis o combinación.

Na₂O + H₂O ⇒ NaOH

Podremos obtener la ecuación balanceada si multiplicamos NaOH por 2.

Na₂O + H₂O ⇒ 2 NaOH

3 0
2 years ago
What is the chemical formula for acetic acid?
liberstina [14]

CH3COOH

Explanation:

hope I helped

if you get the answer right mark me as brainlyst

8 0
2 years ago
How many oxygen atoms are there in the following compound: 2HNO3<br> 3<br> 5<br> 6<br> 12
Fudgin [204]

Answer:

6

Explanation:

There is 2 of the coumpond, in one compound, there are 3 oxygens. But when there are two, it is 6.

7 0
3 years ago
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Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.
madreJ [45]

Answer:

Molar \ solubility=3.12x10^{-5}M

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-

The equilibrium expression is:

Ksp=[Ca^{2+}][F^-]^2

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent x is computed as follows:

3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M

Thus, the molar solubility equals the reaction extent x, therefore:

Molar \ solubility=3.12x10^{-5}M

Regards.

4 0
3 years ago
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