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3 years ago

13

I really need help with this and fast, any you could please answer I would really appreciate it, also please only answer if you

actually do know the answer because people have been taking advantage of me because I’m giving a lot of points and this is due in like 30 min

1 answer:

deff fn [24]3 years ago

5 0

there are 2 types of mixture

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Give 7 and example of how each wave in the EM spectrum in used in our daily lives

bogdanovich [222]

1. it’s use for our food in like microwaves and ovens
2. airplanes are guided by radar waves,
3.the tv is used by electromagnetic waves
4. they are used in heaters,
5.infrared cameras which detect people in the dark
6.Allows airport security to observe the internal contents of objects and luggage using airport scanners
7.doctors use electromagnetic waves in x-rays

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3 years ago

Is classified as a substance <br><br> A. Elements and compound <br><br> B.mixtures

Vika [28.1K]

Pretty sure it’s Mixture if I’m not wrong

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3 years ago

If I add 475mL of water to 75.0mL of a 0.315M NaOH solution, what will the molarity of the diluted solution be?

pantera1 [17]

Answer is: <span>the molarity of the diluted solution 0,043 M.
</span>V(NaOH) = 75 mL ÷ 1000 mL/L = 0,075 L.
c(NaOH) = 0,315 M = 0,315 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,075 L · 0,315 mol/L.
n(NaOH) = 0,023625 mol.
V(solution) = 0,475 L + 0,75 L.
c(solution) = 0,023625 mol ÷ 0,550 L.
c(solution) = 0,043 mol/L.

5 0

3 years ago

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

Dvinal [7]

<u>Answer:</u> The value of $K_c$ for the final reaction is $7.16\times 10^{25}$

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u> $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

<u>Equation 2:</u> $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=\frac{1}{K_1}\times \frac{1}{K_2}$

We are given:

$K_1=9.57\times 10^{-8}$

$K_2=1.46\times 10^{-19}$

Putting values in above equation, we get:

$K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}$

Hence, the value of $K_c$ for the final reaction is $7.16\times 10^{25}$

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3 years ago

ANSWER IN YOUR OWN WORDS!<br> How does the pH of an acid compare with the pH of a neutral solution?

Lady bird [3.3K]

Answer:

A pH of 7 is neutral.A pH of less than 7 is acidic.

Explanation:

5 0

2 years ago

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