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Nuetrik [128]
3 years ago
13

Enter an equation showing how this buffer neutralizes added aqueous acid (HI). Express your answer as a chemical equation. Ident

ify all of the phases in your answer.
Chemistry
1 answer:
andrey2020 [161]3 years ago
5 0

Answer:

H^++NH_3\rightleftharpoons NH_4^+

Explanation:

Hello there!

In this case, since the buffer is not given, we assume it is based off ammonia, it means the ammonia-ammonium buffer, whereas the ammonia is the weak base and the ammonium ion stands for the conjugate acid. In such a way, when adding HI to the solution, the base of the buffer, NH3, reacts with the former to promote the following chemical reaction:

H^++NH_3\rightleftharpoons NH_4^+

Because the HI is totally ionized in solution so the iodide ion becomes an spectator one.

Best regards!

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The units of density are kg/m2. If the density of a liquid is 760.0 kg/m' what is the specific volume? a) 1.316 x 10 m2/kg b) 1.
zlopas [31]

Answer:

The correct answer is: 1.316 . 10⁻³ m³/kg.

Explanation:

The density (ρ) of a substance is the ratio of its <em>mass (m)</em> to its <em>volume (V)</em>. At constant temperature and pressure, its value is constant and it is an intrinsic property of materials. The units of density are kg/m³.

\rho = \frac{m}{V}

The specific volume (ν) of a substance is the ratio of its <em>mass</em> to its <em>volume</em>. We can see that it is the reciprocal of density and an intrinsic property of matter as well. Therefore, the units of specific volume are m³/kg.

\nu = \frac{V}{m}=\frac{1}{\rho }

Given we know the density of the liquid, we can use this relationship to find out its specific volume:

\nu =\frac{1}{\rho }=\frac{1}{760.0kg/m^{3} } =1.316 .10^{-3} m^{3} /kg

6 0
3 years ago
The normal freezing point of a certain liquid
stepladder [879]

Answer : The molal freezing point depression constant of X is 4.12^oC/m

Explanation :  Given,

Mass of urea (solute) = 5.90 g

Mass of X liquid (solvent) = 450.0 g

Molar mass of urea = 60 g/mole

Formula used :  

\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}\times 1000}{\text{Molar mass of urea}\times \text{Mass of X liquid}}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution = -0.5^oC

\Delta T^o = freezing point of liquid X= 0.4^oC

i = Van't Hoff factor = 1  (for non-electrolyte)

K_f = molal freezing point depression constant of X = ?

m = molality

Now put all the given values in this formula, we get

[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}

k_f=4.12^oC/m

Therefore, the molal freezing point depression constant of X is 4.12^oC/m

4 0
3 years ago
A sample of 1.000 g of a compound containing carbon and hydrogen reacts with oxygen at elevated temperature to yield 0.692 g H₂O
ollegr [7]

Answer :

(a) 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

(b) There is no other element present in the compound.

Explanation :

(a) Now we have to determine the masses of C and H in the sample.

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of Carbon and hydrogen respectively.

We are given:

Mass of CO_2=3.381g

Mass of H_2O=0.692g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.381 g of carbon dioxide, \frac{12}{44}\times 3.381=0.922g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.692 g of water, \frac{2}{18}\times 0.692=0.0769g of hydrogen will be contained.

Thus, 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

(b) Now we have to determine the compound contain any other elements or not.

Mass carbon + Mass of hydrogen = 0.922 g + 0.0769 g = 0.999 g ≈ 1 g

This means that there is no other element present in the compound.

3 0
3 years ago
What is the application of chemistry​
vlabodo [156]

Answer:

Chemistry plays an important and useful role towards the development and growth of a number of industries. This includes industries like glass, cement, paper, textile, leather, dye etc. We also see huge applications of chemistry in industries like paints, pigments, petroleum, sugar, plastics, Pharmaceuticals.

3 0
3 years ago
Select all the species that are ions.<br><br> HCl<br> NaCl<br> NH4+<br> H2PO4-<br> CCl4<br> H3O+
nata0808 [166]

Answer:

NH₄⁺

H₂PO₄⁻

H₃O⁺

Explanation:

  • An ion is an atom or molecule with a net electric charge due to the loss or gain of one or more electrons.
  • Ion may be positively charged "cation" or negatively charged "anion".
  • Neutral molecule has a net charge of zero.

<em>So, the species that are ions are: </em>

NH₄⁺

H₂PO₄⁻

H₃O⁺

4 0
4 years ago
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