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Natalija [7]
2 years ago
15

CAN SOMEONE HELP ME HERE PLEASE I ONLY NEED THESE LEFT AND IM CONFUSED-

Chemistry
1 answer:
liubo4ka [24]2 years ago
7 0

Answer:

Pick UpTruck

Explanation:

Because the pick uptruck is bigger and on the oppiste side of the rode

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Fantom [35]
Rub the magnet on the iron an it will cause a stronger force of attraction. Hope this helped!
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3 years ago
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How are water molecules able to weakly bond to one another with hydrogen bonds?
zvonat [6]

C. slight opposite charges

Explanation: opposites attract

8 0
3 years ago
What is the total pressure of three gases, if the partial pressures are: 2.67 mmHg, 45.29 mmHg, and 789.6 mmHg
EastWind [94]

Answer:

The total pressure of three gases is 837.56 mmHg.

Explanation:

The pressure exerted by a particular gas in a mixture is known as its partial pressure. So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

PT = PA + PB

This relationship is due to the assumption that there are no attractive forces between the gases.

In this case, the total pressure can be calculated as:

PT= 2.67 mmHg + 45.69 mmHg + 789.6 mmHg

Solving:

PT= 837.56 mmHg

<em><u>The total pressure of three gases is 837.56 mmHg.</u></em>

7 0
3 years ago
RATE LAW QUESTION !
vivado [14]
In general, we have this rate law express.:

\mathrm{Rate} = k \cdot [A]^x [B]^y
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4)  by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation \Delta \mathrm{Rate} = \Delta [B]^y

3.09 = (3)^y \implies y \approx 1

To this point, \mathrm{Rate} = k \cdot [A]^x [B]^1

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for \Delta \mathrm{Rate} = \Delta [A]^x

4=  (2)^x \implies x = 2

the rate law is

Rate = k·[A]²[B]
6 0
3 years ago
Answer from this picture.
Ganezh [65]
Try photo math it will probably be more helpful
7 0
3 years ago
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