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2 years ago

To stop an object that is in motion on a frictionless surface, you stop applying a force.

Chemistry

1 answer:

madreJ [45]2 years ago

8 0

The correct answer is true, you need to be able to stop it, an object in motion remains in motion unless acted upon

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A star is observed from two positions of Earth in its orbit, in summer and winter. Which of these is the best method to calculat

Fantom [35]

Answer:

D. Calculate the variations in the potions of the star due to movement of Earth in its orbit

6 0

2 years ago

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv

Triss [41]

Answer:

1. $3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2. $5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

Explanation:

The $pH$ value of Aspirin solution = 2.62

$pH=-\log[H^+]$

$[H^+]=10^{-2.62}=0.00240 M$

Moles of s asprin = $\frac{2.00 g}{180 g/mol}=0.01111 mol$

Volume of the solution = 0.600 L

The initial concentration of Aspirin = c = $\frac{0.01111 mol}{0.600 L}=0.0185 M$

$HAs\rightleftharpoons As^-+H^+$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_a=\frac{[As^-][H^+]}{[HAs]}$ :

$K_a=\frac{x\times x }{(c-x)}$

$=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}$

$K_a=3.57\times 10^{-4}$

$3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

The $pH$ value of ethylamine = 11.87

$pH+pOH=14$

$pOH=14-11.87=2.13$

$pOH=-\log[OH^-]$

$[OH^-]=10^{-2.13}=0.00741 M$

The initial concentration of ethylamine = c = 0.100 M

$C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}$ :

$K_b=\frac{x\times x}{(c-x)}$

$=\frac{0.00741\times 0.00741}{(0.100-0.00741)}$

$K_b=5.93\times 10^{-4}$

$5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

3 0

3 years ago

What volume of 0.08892 M HNO3 is required to react completetly with 0.2352 g of potassium hydrogen phosphate?

galina1969 [7]

Answer:

0.0303 Liters

Explanation:

Given:

Mass of the potassium hydrogen phosphate = 0.2352

Molarity of the HNO₃ Solution = 0.08892 M

Now,

From the reaction it can be observed that 1 mol of potassium hydrogen phosphate reacts with 2 mol of HNO₃

The number of moles of 0.2352 g of potassium hydrogen phosphate

= Mass / Molar mass

also,

Molar mass of potassium hydrogen phosphate

= 2 × (39.09) + 1 + 30.97 + 4 × 16 = 174.15 g / mol

Number of moles = 0.2352 / 174.15 = 0.00135 moles

thus,

The number of moles of HNO₃ required for 0.00135 moles

= 2 × 0.00135 mol of HNO₃

= 0.0027 mol of HNO₃

Now,

Molarity = Number of Moles / Volume

thus,

for 0.0027 mol of HNO₃, we have

0.08892 = 0.0027 / Volume

Volume = 0.0303 Liters

8 0

3 years ago

Read 2 more answers

Based on your observations and your understanding about how heat moves (from hot to cold), write a hypothesis that describes whi

Oksi-84 [34.3K]

Answer:

wrapping a boiled water keeps the water hot..cus it's wrapped and no air can go out

3 0

2 years ago

When 3.0 kg of water is warmed from 10 °C to 80 °C, how much heat energy is needed?

Naddik [55]

Answer:

THE HEAT NEEDED TO CHANGE 3KG OF WATER FROM 10 C TO 80 C IS 877.8kJ OR 877,800 J.

Explanation:

Mass = 3.0 kg = 3 * 1000 = 3000 g

Initial temperature = 10 C

Final temperature = 80 C

Change in temperature = 80 - 10 = 70 C

Specific heat of water = 4.18 J/g C

Heat needed = unknown

Heat is the amount of energy in joules needed to change a gram of water by 1 C.

Heat = mass * specific heat * change in temperature

Heat = 3000 g * 4.18 J/g C * 70 C

Heat = 877 800 Joules

Heat = 877.8 kJ.

The heat needed to change 3 kg mass of water from 10 C to 80 C is 877,800 J or 877.8 kJ.

6 0

3 years ago