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Artyom0805 [142]
3 years ago
9

HELP!!!! please omg

Chemistry
1 answer:
OverLord2011 [107]3 years ago
7 0

Answer:

I believe it's A. to reduce air bubbles. Tbh, it's been a while

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You are given 3 unknown solutions with pH value as 6, 8 & 9.5 respectively. Which solution will contain maximum OH ion
Grace [21]
The answer is 9.5 one as it is more basic so it contains more OH ion
Hope it helps : )
6 0
3 years ago
Which beakers in the model contain solid insoluble substances?
ladessa [460]
Although the models are not provided, I was able to find them and the beakers with solid present in them are:
1C
2A
2C
3A
3C

This is determined by the fact that the beakers all have a piece of closely packed substance laying at the bottom. This closely packed lattice is characteristic of solid substances, and the fact that they exist in the solution in the solid states indicates that they are insoluble.
7 0
3 years ago
Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ
JulsSmile [24]

Answer:

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

\frac{1}{2}Br2(g) ⇒ Br(g) ,  Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) ,  Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

   average bond enthalpy is \frac{1134.9}{4} = 283.725 kJ ⋅ mol − 1

4 0
3 years ago
Copper has two naturally occurring isotopes. Cu−63 has a mass of 62.939 amu and relative abundance of 69.17%.
fiasKO [112]
The answer is 64.907 amu.

The atomic mass of an element is the average of the atomic masses of its isotopes. The relative abundance of isotopes must be taken into consideration, therefore:
atomic mass of copper = atomic mass of isotope 1 * abundance 1 + atomic mass of isotope 2 * abundance 2

We know:
atomic mass of copper = 63.546 amu
The atomic mass of isotope 1 is: 62.939 amu
The abundance of isotope 1 is: 69.17% = 0.6917
The atomic mass of isotope 1 is: x
The abundance of isotope 2: 100% - 69.17% = 30.83% = 0.3083

Thus:
63.546 amu = 62.939 amu * 0.6917 + x * 0.3083
63.546 <span>amu = 43.535 amu + 0.3083x
</span>⇒ 63.546 amu - 43.535 amu = 0.3083x
⇒ 20.011 amu = 0.3083x
   ⇒ x = 20.011 amu ÷ 0.3083 = 64.907 amu
7 0
3 years ago
Does anyone know how to do number 15?
vovikov84 [41]

Answer:

yes i do 980

Explanation:

i899 1/2

6 0
3 years ago
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