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3 years ago

When 3000 grams of water is cooled from 80.0 c to 10.0 c, how much energy is released

Chemistry

1 answer:

Rudik [331]3 years ago

4 0

Answer:

877800 Joules of energy is released

Explanation:

mass m = 3000 g

initial temperature = 80 c

Final temperature = 10 c

Change in temperature ΔT = 80 - 10 = 70 c

Heat H = ?

The relationship between these parameters is given by;

H = mcΔT

Where c = specific heat capacity = 4.18 J/gc

H = 3000 * 4.18 * 70 = 877800 Joules

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Consider the reaction:

Flura [38]

Answer:

2 one is the correct one

Explanation:

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3 years ago

A gas has a volume of 590 mL at temperature of -55.0 C. What volume will the gas occupy at 30.0 C show your work

DENIUS [597]

Data:
$V_{initial} = 590\:mL$
$T_{initial} = -55.0^0C$
converting to Kelvin
TK = TC + 273
TK = -55.0 + 273 → TK = 218.0 → $T_{initial} = 218.0\:K$
$V_{final} = ? (in\:milliliters)$
$T_{final} = 30.0^0C$
TK = TC + 273
TK = 30.0 + 273 → TK = 303.0 → $T_{final} = 303.0\:K$

By the first Law of Charles and Gay-Lussac, we have:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$

Solving:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$
$\frac{ 590 }{ 218.0 } = \frac{ V_{f} }{ 303.0 }$
Product of extremes equals product of means:
$218.0* V_{f} = 590*303.0$
$218.0 V_{f} = 178770$
$V_{f} = \frac{178770}{218.0}$
$\boxed{\boxed{V_{f} \approx 820.04\:mL}}\end{array}}\qquad\quad\checkmark$

7 0

3 years ago

HELP<br> FREE BRAINLIEST

liberstina [14]

The universe comes into existence is first
The first neutral atoms form is second
The universe begins expanding is third
Gases form that will later go to shape stars and galaxies is fourth
Atomic nuclei form is last

I'm almost certain that is correct. Do not take my word for this.

6 0

3 years ago

Read 2 more answers

What is the ROUNDED atomic number for this elements?

denis-greek [22]

Sodium/Atomic number

11

Gold/Atomic number

79
Potassium/Atomic number

19

Silicon/Atomic number

14

8 0

3 years ago

If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×

Elodia [21]

Answer:

2.30 × 10⁻⁶ M

Explanation:

Step 1: Given data

Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M

Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³

Step 2: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂

We will use the following expression.

Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²

[OH⁻] = 2.30 × 10⁻⁶ M

3 0

4 years ago