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emmasim [6.3K]
3 years ago
14

Convert 256.3 g sodium carbonate to formula units

Chemistry
1 answer:
nalin [4]3 years ago
7 0

Answer:

1.46 X 10^24 fu Na2CO3

Explanation:

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How many moles of phosphorus are in a sample of 40.4 grams of phosphorus I NEED HELP FAST
steposvetlana [31]

Answer:

1.30 mol P

Explanation:

take what is given to you (40.4g). now divide it by the molar mass of the asked element (phosphorus = 30.97g/mol)

40.4g P / 30.97g/mol P = 1.30 mol P

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3 years ago
It requires two sticks of butter to make a batch of 20 cookies. How much butter will it take to make 150 cookies?
liq [111]

Answer:

It would take 15 sticks of butter

Explanation:

2 divided by 20 is 0.1

0.1 is how much butter for 1 so now we multiply that by 150

0.1 x 150 = 15

Therefore, you would need 15 sticks of butter for 150 cookies

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3 years ago
The spectral lines energetic object known as quasars
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Quasars inhabit the centers of active galaxies and are among the most luminous, powerful, and energetic objects known in the universe, emitting up to a thousand times the energy output of the Milky Way, which contains 200–400 billion stars.

Explanation:

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3 years ago
What is the fourth quantum number of the 3p electron in aluminum,
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8 0
3 years ago
A chunk of tin weighing 18.5 grams and originally at 97.38 °C is dropped into an insulated cup containing 75.7 grams of water at
weqwewe [10]

Answer:

22.44°C will be the final temperature of the water.

Explanation:

Heat lost by tin will be equal to heat gained by the water

-Q_1=Q_2

Mass of tin = m_1=18.5 g

Specific heat capacity of tin = c_1=0.21 J/g^oC

Initial temperature of the tin = T_1=97.38^oC

Final temperature = T_2=T

Q_1=m_1c_1\times (T-T_1)

Mass of water= m_2=75.7 g

Specific heat capacity of water= c_2=4.184 J/g^oC

Initial temperature of the water = T_3=21.52^oC

Final temperature of water = T_2=T

Q_2=m_2c_2\times (T-T_3)

-Q_1=Q_2

-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)

On substituting all values:

-(18.5 g\times 0.21 J/g^oC\times (T-97.38^oC))=75.7 g\times 4.184 J/g^oC\times (T-21.52 ^oC)

we get, T = 22.44°C

22.44°C will be the final temperature of the water.

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4 years ago
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