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2 years ago

Rewrite this measurement with a simpler unit, if possible. 8.3 kg•m^2/ s^2

Chemistry

1 answer:

Feliz [49]2 years ago

7 0

Answer: 8.3 J

Explanation:

We have the following measurement:

$8.3 \frac{kg m^{2}}{s^{2}}$

Rearranging the units:

$8.3 \frac{kg m}{s^{2}}m$

Since 1 Newton is $1 N=1 \frac{kg m}{s^{2}}$ :

$8.3 Nm$

Since 1 Joule is $1 J=1 Nm$ :

$8.3 J$ This is the simplest form possible

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Answer:

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Explanation:

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3 years ago

Compare the properties of elements in groups 17 and group 18. Make sure to cover the following points:

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The answer for this question is b

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2 years ago

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Consider the following reaction where Kc = 154 at 298 K.2NO(g) + Br2(g) 2NOBr(g)A reaction mixture was found to contain 4.64×10-

IgorLugansk [536]

Answer:

The reaction is not at equilibrium and reaction must run in forward direction.

Explanation:

At the given interval, concentration of NO = $\frac{4.64\times 10^{-2}}{1}M=4.64\times 10^{-2}M$

Concentration of $Br_{2}$ = $\frac{4.56\times 10^{-2}}{1}M=4.56\times 10^{-2}M$

Concentration of NOBr = $\frac{0.102}{1}M=0.102M$

Reaction quotient, $Q_{c}$ , for this reaction = $\frac{[NOBr]^{2}}{[NO]^{2}[Br_{2}]}$

species inside third bracket represents concentrations at the given interval.

So, $Q_{c}=\frac{(0.102)^{2}}{(4.64\times 10^{-2})^{2}\times (4.56\times 10^{-2})}=106$

So, the reaction is not at equilibrium.

As $Q_{c}< K_{c}$ therefore reaction must run in forward direction to increase $Q_{c}$ and make it equal to $K_{c}$ .

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2 years ago

Please help :)

melamori03 [73]

The driving thrust of the car produced by the engine is the main forward force.

The main opposing forces are air resistance (from the wind) and friction (between the tyres and the road)

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2 years ago

Calculate the molecular weight of a substance. In which the solution of this substance in the water has a concentration of 7 per

Eduardwww [97]

Answer : The molecular weight of a substance is 157.3 g/mol

Explanation :

As we are given that 7 % by weight that means 7 grams of solute present in 100 grams of solution.

Mass of solute = 7 g

Mass of solution = 100 g

Mass of solvent = 100 - 7 = 93 g

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T_f^o-T_f=k_f\times\frac{\text{Mass of substance(solute)}\times 1000}{\text{Molar mass of substance(solute)}\times \text{Mass of water(solvent)}}$