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Alborosie
3 years ago
12

Is this chemical equation balanced?

Chemistry
2 answers:
cluponka [151]3 years ago
5 0

Answer:

Yes

Explanation:

Lana71 [14]3 years ago
4 0
Not at all

C6H12 + O2 — 6CO2 + 6H2O

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Help, due tonight (2/19/2021) at 11:59pm!
cricket20 [7]

Answer:

283549.68 J

Explanation:

From the question given above, the following data were obtained:

Mass (M) = 2.7 Kg

Initial Temperature (T₁) = 5.4 °C

Final temperature (T₂) = 30.5 °C

Specific heat capacity (C) = 4184 J/KgºC

Heat (Q) =?

The amount of heat energy needed can be obtained as follow:

Q = MC(T₂ – T₁)

Q = 2.7 × 4184 × (30.5 – 5.4)

Q = 11296.8 × 25.1

Q = 283549.68 J

Thus, the heat energy needed is 283549.68 J

8 0
3 years ago
The graph shows the distribution of energy in the particles of two gas samples at different temperatures, T1 and T2. A, B, and C
scoray [572]

Answer: More information needed. Where's the graph??????????????

Explanation:

3 0
3 years ago
For a substance to change phases, the amount of internal energy must change. Water exists in three phases: liquid, solid (ice),
vfiekz [6]

Answer:

Solid, liquid, gas

Explanation:

add thermal energy to solid and it becomes liquid, add thermal energy to liquid it becomes gas

3 0
3 years ago
Read 2 more answers
Consider the following reactions:
Advocard [28]
What’s the question cuz I’m confused lol
5 0
3 years ago
) A technique once used by geologists to measure the density of a mineral is to mix two dense liquids in such proportions that t
sergeinik [125]

Hey there!

It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.

In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:

m_{calcite}=15.4448g-12.4631g=2.9817g

Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:

V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL

Thus, the density of the calcite sample will be:

\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL

This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL

Learn more:

  • brainly.com/question/12001845
  • brainly.com/question/11242138

Regards!

6 0
2 years ago
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