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3 years ago

NEEDS TO BE CORRECT EXPLAIN!!!!! I'M GIVING 50 POINTS!!!!!!!

2 answers:

7 0

In an ammonia solution (NH4OH):
The number of nitrogen (N) atoms is 1
The number of hydrogen (H) atoms is 5 (the H4 AND the H)
The number of oxygen (O) atoms is 1
So, there are 3 different TYPES of element, but 7 atoms altogether in this compound.
However if you are referring to ammonia on its own, there are 4 atoms as ammonia is written as NH3, so has 1 nitrogen atom and 3 hydrogen atoms

Minchanka [31]3 years ago

5 0

Hello there,

Ammonia will have a total of 4 atoms.

Hope I helped :)

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Can cells have different types of openings?

Tcecarenko [31]

Answer:

Yes they can.

Explanation:

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4 years ago

Type of force assessment for midd school

navik [9.2K]

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3 0

3 years ago

Consider the following reaction where K. = 154 at 298 K: 2NO(g) + Brz(9) 2NOBr(g) A reaction mixture was found to contain 2.69x1

bekas [8.4K]

Explanation:

$2NO(g) + Br_2(g)\rightleftharpoons 2NOBr(g)$

Equilibrium constant of reaction = $K=154$

Concentration of NO = $[NO]=\frac{2.69\times 10^{-2} mol}{1 L}=2.69\times 10^{-2} M$

Concentration of bromine gas = $[Br_2]=\frac{3.85\times 10^{-2} mol}{1 L}=3.85\times 10^{-2} M$

Concentration of NOBr gas = $[Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M$

The reaction quotient is given as:

$Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}$

$Q=328.06$

$Q>K$

The reaction will go in backward direction in order to achieve an equilibrium state.

1. In order to reach equilibrium NOBr (g) must be produced. False

2. In order to reach equilibrium K must decrease. False

3. In order to reach equilibrium NO must be produced. True

4. Q. is less than K . False

5. The reaction is at equilibrium. No further reaction will occur. False

8 0

3 years ago

What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?

Ymorist [56]

Answer:

$\% diss = 50\%$

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

$HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}$

Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:

$Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}$

Thus, it is possible to find x given the pH as shown below:

$x=10^{-pH}=10^{-4.74}=1.82x10^{-5}M$

So that we can calculate the initial concentration of the acid:

$\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\$

$[HA]_0=3.64x10^{-5}M$

Therefore, the percent dissociation turns out to be:

$\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%$

Best regards!

6 0

3 years ago

WILL MARK BRAINLIEST

levacccp [35]

Okay so i think to calculate volume change you use the ideal gas law . this is pressure x volume = amount of substance x ideal gas constant x temperature. i honestly don’t know where to go from there but i hope this helped a bit :(

8 0

3 years ago