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julsineya [31]
4 years ago
13

If 20 grams of aluminum react with 200 grams of bromide to form aluminum bromide and no aluminum is left after the reaction, but

23 grams of bromine remained unreacted. how many grams of aluminum bromide were formed
Chemistry
1 answer:
Varvara68 [4.7K]4 years ago
8 0
The answer is: 197 grams

Why?

All 20 grams of aluminum was used to form aluminum bromide. Only 177 grams of bromide were used to make aluminum bromide, since 23 grams were left over.

20 grams + 177 grams = 197 grams, so the answer is 197 grams.
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At what temperature does the motion of the particles of an ideal gas cease?
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Answer:The temperature at which the motion of particles theoretically ceases is 00A or absolute zero, Absolute 00 is the definition of 00 on the Kelvin and Rankine temperature scales.

Explanation:

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2 years ago
Which of the following is part of the molarity equation?
Zinaida [17]
Answer:

Moles of Solution
4 0
3 years ago
How many moles of Carbon are in 3.06 g of Carbon
natta225 [31]

Answer:

\boxed {\boxed {\sf 0.255 \ mol \ C }}

Explanation:

If we want to convert from grams to moles, the molar mass is used. This is the mass of 1 mole. They are found on the Periodic Table as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

Look up the molar mass of carbon.

  • Carbon (C): 12.011 g/mol

Set up a ratio using the molar mass.

\frac {12.011 \ g \ C}{ 1 \ mol \ C}

Since we are converting 3.06 grams to moles, we multiply by that value.

3.06 \ g \ C*\frac {12.011 \ g \ C}{ 1 \ mol \ C}

Flip the ratio. This way, the ratio is still equivalent, but the units of grams of carbon cancel.

3.06 \ g \ C* \frac{1 \ mol \ C}{12.011 \ g\ C}                      

3.06 * \frac{1 \ mol \ C}{12.011 }    

\frac {3.06}{12.011 } \ mol \ C                                

0.25476646 \ mol \ C

The original measurement of grams (3.06) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

  • 0.25476646

The 7 in the ten-thousandth place tells us to round the 4 up to a 5.

0.255 \ mol \ C

3.06 grams of carbon is approximately <u>0.255 moles of carbon.</u>

3 0
3 years ago
How many grams of water at 25 C must be mixed in an insulated container with 250 grams of water at 93 C if the final temperature
nydimaria [60]

Answer:

Explanation

Comment

Normally you would just use m*c*delta T. But the c is the same in both cases, so you need just use m*deltaT

Givens

25 degrees

m = ?

T = 25

t = final temp = 75

93 degrees

m = 250

T = 93

t = final temp = 75

Solution

(t - 25)*m = (93 - t) * 250                   Remove the brackets

75*m - 25m = 93*250 -  250*75      Combine the right

m(75 - 25) = 23250 - 250*75

50 m = 23250 - 18750                      Combine

50 m = 4500                                      Divide both sides by 50

50*m/50 = 4500/50

m = 90

Answer: 90 grams are needed.

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Warm and cold fronts are formed by the movement of different air masses.
vekshin1
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