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2 years ago

How much heat energy is needed to heat 300g of water from 10 degrees Celsius to 50 degrees Celsius

Chemistry

1 answer:

elixir [45]2 years ago

8 0

Answer:

There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

Explanation:

<u>Step 1: </u>Data given

mass of water = 300 grams

initial temperature = 10°C

final temperature = 50°C

Temperature rise = 50 °C - 10 °C = 40 °C

Specific heat capacity of water = 4.184 J/g °C

<u>Step 2:</u> Calculate the heat

Q = m*c*ΔT

Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)

Q = 50208 Joule = 50.2 kJ

There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

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If distance, d, is proportional to time, t, and at time t =2.00 seconds, the distance is 6.20 m, find the distance when t = 3.50

Allushta [10]

At 3.5s the distance would be 10.85

Find 1/4th of 2.00 and add that to 6.2

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3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

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3 years ago

If an ideal gas has a pressure of 2.97 atm, a temperature of 449 K, and has a volume of 58.35 L, how many miles of gas are in th

soldi70 [24.7K]

N = ?

T = 449 K

V = 58.35 L

P =2.97

R = 0.082

Use the clapeyron equation:

P x V = n x R x T

2.97 x 58.35 = n x 0.082 x 449

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hope this helps!

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3 years ago

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Answer:

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