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ikadub [295]
3 years ago
13

Emits huge quantities of light and heat due to the nuclear

Chemistry
1 answer:
kifflom [539]3 years ago
8 0
Black hole I think that is right
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C i took the test and got it correct

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2 years ago
How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?
Montano1993 [528]

Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

so you have (144.3x1.4/100) g glucose= 2.0202 grams

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3 years ago
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Burning gasoline is an _____________ reaction.
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3 0
3 years ago
A sample of gas occupies 40.0 mL at -123°C. What volume does the sample occupy at 27°C if the pressure is held constant? V1T2 =
Brums [2.3K]

Answer:

80mL

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 40mL

Initial temperature (T1) = –123°C

Final temperature (T2) = 27°C

Final volume (V2) =..?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

T(K) = T(°C) + 273

Initial temperature (T1) = –123°C =

–123°C + 273 = 150K

Final temperature (T2) = 27°C = 27°C + 273 = 300K

Step 3:

Determination of the final volume.

This can be obtained as follow:

V1/T1 = V2/T2

Initial Volume (V1) = 40mL

Initial temperature (T1) = 150K

Final temperature (T2) = 300

Final volume (V2) =..?

V1/T1 = V2 /T2

40/150 = V2 /300

Cross multiply

150 x V2 = 40 x 300

Divide both side by 150

V2 = (40 x 300) /150

V2 = 80mL

Therefore, the new volume of the gas is 80mL

6 0
3 years ago
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