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3 years ago

15

In which of the following situations is the greatest amount of work accomplished?

2 answers:

Dahasolnce [82]3 years ago

7 0

<span>work = force x distances </span>

<span>A - moving 2 newton's up 0.6 meter = 1.2 joules
B - moving 4 newton's up 0.6 meter = 2.4 joules
C - moving 6 newton's up 0.3 meter = 1.8 joules
D - moving 9 newton's up 0.3 meter = 2.7 joules
The greatest amount of work is in example D.
</span>

<span>
D is your answer. </span>

UNO [17]3 years ago

6 0

The answer is D “A boy lifts a 9-newton box 3.0 meters”

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_________ energy is to the fission and fusion within an atom as ________ energy is to the fission of atomic compounds.

xeze [42]

I would say C bc I don’t rlly think it’s A or B

7 0

3 years ago

Read 2 more answers

What kind of changes occur between atoms when substances undergo chemical reactions?

KatRina [158]

The atoms undergo chemical changes.

New substances may be a product of the reaction.

8 0

3 years ago

Be sure to answer all parts.The equilibrium constant, Kc, for the formation of nitrosyl chloride from nitric oxide and chlorine,

djverab [1.8K]

<u>Answer:</u> The reaction proceeds in the forward direction

<u>Explanation:</u>

For the given chemical equation:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

Relation of $K_p\text{ with }K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta n_g}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = $6.5\times 10^4$

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $35^oC=[35+273]K=308K$

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$K_p=6.5\times 10^4\times (0.0821\times 500)^{-1}\\\\K_p=1583.43$

$K_p$ is the constant of a certain reaction at equilibrium while $Q_p$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

The expression of $Q_p$ for above equation follows:

$Q_p=\frac{(p_{NOCl})^2}{p_{Cl_2}\times (p_{NO})^2}$

We are given:

$p_{NOCl}=1.76atm$

$p_{NO}=1.01atm$

$p_{Cl_2}=0.42atm$

Putting values in above equation, we get:

$Q_p=\frac{(1.76)^2}{0.42\times (1.01)^2}=7.23$

We are given:

$K_p=1583.43$

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium

As, $K_p>Q_p$ , the reaction will be favoring product side.

Hence, the reaction proceeds in the forward direction

4 0

3 years ago

VLD [36.1K]

Answer:

<h3>The answer is 8.29 %</h3>

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\% \\$

From the question

actual density = 19.30g/L

error = 20.9 - 19.3 = 1.6

We have

$p(\%) = \frac{1.6}{19.3} \times 100 \\ = 8.290155440...$

We have the final answer as

<h3>8.29 %</h3>

Hope this helps you

5 0

4 years ago

A solution of acetic acid and water contains 205.0 g L-1 of acetic acid and 820.0 g L-1 of water. Compute the density of the sol

LekaFEV [45]

Answer:

$\rho_t=1025000 gmL^{-1}$

Explanation:

From the question we are told that:

Density of acetic acid $\rho_a=205 gL^{-1}$

Density of Water $\rho_w=820 gL^{-1}$

Generally the equation for Solution Density is mathematically given by

$\rho_t= \rho_w+\rho_a$

$\rho_t=205+820$

$\rho_t=1025 gL^{-1}$

$\rho_t=1025000 gmL^{-1}$

6 0

3 years ago