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NISA [10]
2 years ago
7

How many donuts are in a mole of donuts ?​

Chemistry
2 answers:
zhannawk [14.2K]2 years ago
7 0

6.02×10²³ donuts is how much donuts are in a mole.

pychu [463]2 years ago
5 0

Whats the donar mass

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Suppose that 0.48 g of water at 25 ∘ C condenses on the surface of a 55- g block of aluminum that is initially at 25 ∘ C . If th
mamaluj [8]

Answer:

49^oC

Explanation:

At 25^oC, the heat of vaporization of water is given by:

\Delta H^o_{vap} = 43988 J/mol\cdot \frac{1 mol}{18.016 g} = 2441.6 J/g

The water here condenses and gives off heat given by the product between its mass and the heat of vaporization:

Q_1 = \Delta H^o_{vap} m_w

The block of aluminum absorbs heat given by the product of its specific heat capacity, mass and the change in temperature:

Q_2 = c_{Al}m_{Al}(t_f - t_i)

According to the law of energy conservation, the heat lost is equal to the heat gained:

Q_1 = Q_2 or:

\Delta H^o_{vap} m_w = c_{Al}m_{Al}(t_f - t_i)

Rearrange for the final temperature:

\Delta H^o_{vap} m_w = c_{Al}m_{Al}t_f - c_{Al}m_{Al}t_i

We obtain:

\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i = c_{Al}m_{Al}t_f

Then:

t_f = \frac{\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i}{c_{Al}m_{Al}} = \frac{2441.6 J/g\cdot 0.48 g + 0.903 \frac{J}{g^oC}\cdot 55 g\cdot 25^oC}{0.903 \frac{J}{g^oC}\cdot 55 g} = 49^oC

6 0
3 years ago
Question 6 (1 point)
Mnenie [13.5K]

Answer:

The correct option is;

d 4400

Explanation:

The given parameters are;

The mass of the ice = 55 g

The Heat of Fusion = 80 cal/g

The Heat of Vaporization = 540 cal/g

The specific heat capacity of water = 1 cal/g

The heat required to melt a given mass of ice = The Heat of Fusion × The mass of the ice

The heat required to melt the 55 g mass of ice = 540 cal/g × 55 g = 29700 cal

The heat required to raise the temperature of a given mass ice (water) = The mass of the ice (water) × The specific heat capacity of the ice (water) × The temperature change

The heat required to raise the temperature of the ice from 0°C to 100°C = 55 × 1 × (100 - 0) = 5,500 cal

The heat required to vaporize a given mass of ice = The Heat of Vaporization × The mass of the ice

The heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal

The total heat required to boil 55 g of ice = 29700 cal + 5,500 cal + 4,400 cal = 39,600 cal

However, we note that the heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal.

The heat required to vaporize the 55 g mass of ice at 100°C = 4,400 cal

3 0
2 years ago
What is the reason for heat transfer from one substance to another?
Yuliya22 [10]
Hey man its difference in temperature
3 0
3 years ago
Read 2 more answers
Help question 9 is due today also :)
devlian [24]

Answer:

Car 3 with a net force of 12N

Explanation:

The formula is F=MA

Hope this helps friend

8 0
3 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is
Pie

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

Step 5: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

Step 6: Calculate mass ethane

Mass ethane = moles ethane * molar mass ethane

Mass ethane = 0.268 moles * 30.07 g/mol

Mass ethane = 8.06 grams

There will remain 8.06 grams of ethane

7 0
3 years ago
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