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2 years ago

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A family leaves for summer vacation by driving on the highway. The car's tires start the trip with a pressure of 3.8atm at a tem

perature of 19oC and a volume of 1.7L. What is the pressure of the tires after driving, when the temperature within the tire increases to 105oC?

1 answer:

goldenfox [79]2 years ago

4 0

Answer: The pressure of the tires after driving is 4.9atm

Explanation: Please see the attachments below

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Jeff has 10 grams of water and 10 grams of vegetable oil in separate containers. Both liquids have a temperature of 24°C. Jeff h

Anna11 [10]

I think the answer would be that vegetable oil is a better conductor of heat than water.

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A hydrogen atom is removed from the first carbon atom of a butane molecule and is replaced by a hydroxyl group. draw the new mol

Leno4ka [110]

The new molecule that is formed is an alcohol and is known as butanol. It has a chemical formula CH3CH2CH2CH2OH or C4H9<span>OH</span>. A molecule that has a hydroxyl functional group is an alcohol. It has isomers namely tert-butanol, 2-butanol and isobutanol.

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2 years ago

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

Now put all the given values in the above formula, we get:

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

Now put all the given values in the above formula, we get:

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

4 0

2 years ago

1. A crane lifts a 75kg mass a height of 8 m. Calculate the gravitational potential energy

koban [17]

Answer:

GP.E = 5880 j

Explanation:

Given data:

Mass = 75 kg

height = 8 m

Potential energy = ?

Solution:

The formula for gravitational potential energy is

GPE = mgh

m = mass in kilogram

g = acceleration due to gravity

h = height in meter above the ground

Formula:

GP.E = mgh

Now we will put the values in formula.

g = 9.8 m/s²

GP.E = 75 Kg × 9.8 m/s²× 8 m

GP.E = 5880 Kg.m²/s²

Kg.m²/s² = j

GP.E = 5880 j

6 0

2 years ago

An archeologist finds a 1.62 kg goblet that she believes to be made of pure gold. She determines that the volume of the goblet i

katovenus [111]

The answer is

<span>The density (D) is quotient of mass (m) and volume (V):
</span> $D= \frac{m}{V}$
The unit is g/cm³

It is given:
m = 1.62 kg = 1620 g
V = 205 mL = 205 cm³
D = ?

Thus:
$D= \frac{m}{V} = \frac{1620g}{205cm ^{3} } = 7.90 g/cm ^{3}$

The density of the goblet is 7.90 g/cm³.

5 0

2 years ago