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2 years ago

If you increase the force on an object, its acceleration increases. True or false

Chemistry

1 answer:

lions [1.4K]2 years ago

8 0

Answer:

true

Explanation:

from Newton's second law of motion ,we can understand that acceleration of a body is directly proportional to the force applied on it and inversely proportional to its mass ,so as the force applied increase,the acceleration also increases

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When you apply 1000 joules of energy to 50 grams of water its temperature changes to 30 degrees . What was the initial temperatu

expeople1 [14]

Answer:

25.2°C

Explanation:

Given parameters:

Energy applied to the water = 1000J

Mass of water = 50g

Final temperature = 30°C

Unknown:

Initial temperature = ?

Solution:

To solve this problem, we use the expression below:

H = m c Ф

H is the energy absorbed

m is the mass

c is the specific heat capacity

Ф is the change in temperature

1000 = 50 x 4.184 x (30 - initial temperature )

1000 = 209.2(30 - initial temperature)

4.78 = 30 - initial temperature

4.78 - 30 = - initial temperature

Initial temperature = 25.2°C

7 0

2 years ago

What is the answers?

ra1l [238]

the numbers are going to be small so like a power but its at the bottom

NH3, H2O2, NHO2

5 0

3 years ago

The gas in the piston is at constant temperature. A student increases the pressure on the piston from 2 atm to 3 atm. The observ

Vesna [10]

Answer: row 1, the volume decreases when the pressure increased

Explanation:

7 0

3 years ago

If a mineral contains only atoms of oxygen and metal, it is most likely

wariber [46]

Answer: If a mineral contains only atoms of oxygen and metal, it is most likely an oxide mineral.

Explanation:

5 0

3 years ago

Read 2 more answers

7. NH2CO2NH4(s) when heated to 450 K undergoes the following reaction to produce a system which reaches equilibrium: NH2CO2NH4(s

Taya2010 [7]

Answer:

Value of equilibrium constant is 0.0888

Explanation:

Both $NH_{3}$ and $CO_{2}$ are gaseous. Hence equilibrium constant depends upon partial pressures of $NH_{3}$ and $CO_{2}$ .

Initially no $NH_{3}$ and $CO_{2}$ were present.

Hence mole fraction of $NH_{3}$ and $CO_{2}$ at equilibrium can be calculated from coefficient of $NH_{3}$ and $CO_{2}$ in balanced equation.

Mole fraction of $NH_{3}$ = (number of moles of $NH_{3}$ )/(total number of moles of $NH_{3}$ and $CO_{2}$ ) = $\frac{2moles}{(2+1)moles}=\frac{2}{3}$

Mole fraction of $CO_{2}$ = (number of moles of $CO_{2}$ )/(total number of moles of $NH_{3}$ and $CO_{2}$ ) = $\frac{1moles}{(2+1)moles}=\frac{1}{3}$

Let's assume both $CO_{2}$ and $NH_{3}$ behaves ideally.

Therefore partial pressure of $NH_{3}$ , $P_{NH_{3}}= x_{NH_{3}}.P_{total}$ and P_{CO_{2}}= x_{CO_{2}}.P_{total}

Where x represents mole fraction

So, $P_{NH_{3}}=\frac{2}{3}\times 0.843atm=0.562atm$

$P_{CO_{2}}=\frac{1}{3}\times 0.843atm=0.281atm$

So, $K_{p}=P_{NH_{3}}^{2}.P_{CO_{2}}=(0.562)^{2}\times 0.281=0.0888$

4 0

3 years ago