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11111nata11111 [884]
3 years ago
14

How much heat energy in megajoules is neede to convert 7kilograms of ice at negative 9 degrees to water at 0 degrees celsius

Chemistry
1 answer:
den301095 [7]3 years ago
8 0

Heat_1: Get the ice to 0 degrees

Convert 7 kg to grams

7 kg [1000 grams / 1 kg] = 7000 grams

Heat needed to get the the ice from - 9 to 0

deltat = 0 - -9 = 9 degrees

m = 7000 grams

c = 2.1 joules/gram

Heat_1 = m*c*deltat

Heat_1 = 7000 * 2.1 * 9

Heat_1 = 132,300 joules

Heat_2: Melt the ice.

There is no temperature change. The formula is 333 j/gram

Formula: H = mass * constant

H = 7000 g * 333 J / gram

H = 2331000 joules

Heat_3: Total amount of Joules needed.

2331000 + 132300 = 2 463 300 joules

Convert to Megajoules

2 463 300 joules * 1 megajoule / 1000000 = 2.63 megajoules.

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sergij07 [2.7K]

Answer:

to simplify the number by using fewer digits

Explanation:

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3 years ago
If 5 L of butane is reacted what volume of carbon dioxide is produced ILL GIVE BRAINLIEST
Len [333]

Answer: First, here is the balanced reaction:  2C4H10  +  13O2  ===>  8CO2  +  10H2O.

This says for every mole of butane burned 4 moles of CO2 are produced, in other words a 2:1 ratio.

Next, let's determine how many moles of butane are burned.  This is obtained by

5.50 g / 58.1 g/mole  =  0.0947 moles butane.  As CO2 is produced in a 2:1 ratio, the # moles of CO2 produced is 2 x 0.0947  =  0.1894 moles CO2.

Now we need to figure out the volume.  This depends on the temperature and pressure of the CO2 which is not given, so we will assume standard conditions:  273 K and 1 atmosphere.

We now use the ideal gas law PV = nRT, or V =nRT/P, where n is the # of moles of CO2, T the absolute temperature, R the gas constant (0.082 L-atm/mole degree), and P the pressure in atmospheres ( 1 atm).

V = 0.1894 x 0.082 x 273.0 / 1  =  4.24 Liters.

Explanation:

8 0
3 years ago
Gold forms a substitutional solid solution with silver. Calculate the number of gold atoms per cubic centimeter (in atoms/cm3) f
Leokris [45]

Answer:

Gold: 1.1 x 10²² atoms/cm³

Silver: 4.8 x 10²² atoms/cm³

Explanation:

100 g of the alloy will have 29 g of Au and 71 g of Ag.

19.32 g Au ____ 1 cm³

29 g Au ______  x

x = 1.5 cm³

10.49 g Ag ____ 1 cm³

71 g Ag _______   y

y = 6.8 cm³

The total volume of 100g of the alloy is x+y = 8.3 cm³.

Gold:

196.97 g Au____ 6.022 x 10²³ atoms Au

29 g Au _______ w

w = 8.9 x 10²² atoms Au

8.9 x 10²² atoms Au ____ 8.3 cm³

           A                    ____ 1 cm³

A = 1.1 x 10²² atoms Au

Silver:

107.87 g Ag____ 6.022 x 10²³ atoms Ag

71 g Ag _______ w

w = 4.0 x 10²³ atoms Ag

4.0 x 10²³ atoms Ag ____ 8.3 cm³

          B                    ____ 1 cm³

B = 4.8 x 10²² atoms Ag

6 0
3 years ago
Please help me in my TEST
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Answer:

D) Please look below at the cart to cheak your answer!

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4 0
4 years ago
How many formula units are there in 20.6 moles of Al(NO3)3?
Ket [755]

Answer:

How many moles Al(NO3)3 in 1 grams? The answer is 0.0046949186022713.

We assume you are converting between moles Al(NO3)3 and gram.

You can view more details on each measurement unit:

molecular weight of Al(NO3)3 or grams

This compound is also known as Aluminium Nitrate.

The SI base unit for amount of substance is the mole.

1 mole is equal to 1 moles Al(NO3)3, or 212.996238 grams.

Note that rounding errors may occur, so always check the results.

Use this page to learn how to convert between moles Al(NO3)3 and gram.

Type in your own numbers in the form to convert the units!

Explanation:

5 0
3 years ago
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