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GenaCL600 [577]
2 years ago
10

Which type of galaxy contains mostly new stars because it has lots of dust and gas

Chemistry
1 answer:
uysha [10]2 years ago
8 0
Spiral galaxies .........
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An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the
victus00 [196]

<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M

  • To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

pOH=14-8.08=5.92

  • To calculate pOH of the solution, we use the equation:

pOH=-\log[OH^-]

Putting values in above equation, we get:

5.92=-\log[OH^-]

[OH^-]=10^{-5.92}=1.202\times 10^{-6}M

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of X^- ions follows:

                    X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b

<u>Initial:</u>              0.1

<u>At eqllm:</u>        0.1-x                           x              x

Concentration of OH^-=x=1.202\times 10^{-6}M

The expression of K_b for above equation follows:

K_b=\frac{[OH^-][HX]}{[X^-]}

Putting values in above expression, we get:

K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M

  • To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:  

K_w=K_b\times K_a

where,

K_w = Ionic product of water = 10^{-14}

K_a = Acid dissociation constant

K_b = Base dissociation constant = 1.445\times 10^{-11}

Putting values in above equation, we get:

10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}

We know that:

K_a\text{ for HF}=6.8\times 10^{-6}

K_a\text{ for HCl}=1.3\times 10^{6}

K_a\text{ for HClO}=3.0\times 10^{-8}

So, the calculated K_a is approximately equal to the K_a of HF

Hence, the unknown salt is NaF

6 0
3 years ago
C₇H₆O₂ + O₂ --&gt;CO₂ +H₂O Find the chemical reaction
lions [1.4K]

Answer:

Air

Explanation:

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2 years ago
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Two processes are described below:
UNO [17]

Answer:

i would say D i just did this but i kinda forgot so sorry if im wrong or A

Explanation:

4 0
3 years ago
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Antoine put some metal into a container with air. He closed the container. He measured the weight of the closed container with t
Mandarinka [93]

Answer:

the same

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due to the law of conservation of mass, the mass will not change

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Identify the weak diprotic acid. identify the weak diprotic acid. h2so4 hcooh
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Among formic acid (HCOOH ) and sulfuric acid (H₂SO₄), formic acid is the weak acid. Acidic strength of any acid is the tendency of that acid to loose proton. Among these two acids formic acid has a pKa value of 3.74 greater than that of sulfuric acid i.e. -10. Remember! Greater the pKa value of acid weaker is that acid and vice versa. Below I have drawn the Ionization of both acids to corresponding conjugate bases and protons. The structures below with charges are drawn in order to explain the reason for strength. As it is seen in charged structure of formic acid, there is one positive charge on carbon next to oxygen carrying proton. The electron density is shifted toward carbon as it is electron deficient and demands more electron hence, attracting electron density from oxygen and making the oxygen hydrogen bond more polar. While, in case of sulfuric acid it is depicted that Sulfur attached to oxygen containing proton has 2+ charge, means more electron deficient as compared to carbon of formic acid, hence, more electron demanding and strongly attracting electrons from oxygen and making the oxygen hydrogen bond very polar and highly ionizable.

7 0
3 years ago
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