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Archy [21]
3 years ago
10

How much heat is released when 10.0 grams of a substance cools 15 degrees? The specific

Chemistry
1 answer:
VLD [36.1K]3 years ago
4 0

Answer:

Q = 525 J

Explanation:

Given that,

Mass, m = 10 grams

The change in temperature, \Delta T=15^{\circ} C

The specific heat of the substance is 3.5 J/g °C

We need to find the amount of heat released in the process. It can be given by the formula as follow :

Q=mc\Delta T\\\\Q=10\ g\times 3.5\ J/g^{\circ} C\times 15^{\circ} C\\Q=525\ J

Hence, 525 J of heat is released.

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. I am described as being conductive, malleable, shiny, and reactive with HCl. I am one of the most reactive elements. I turn bl
mylen [45]

Answer:

Potassium

Explanation:

Most reactive metal in the reactivity series

8 0
3 years ago
The equilibrium constant for the gas phase reaction
vampirchik [111]

only reactants are present

If equllibrium constant is greater than 10^3 then product predominates

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8 0
3 years ago
For the generic equilibrium HA(aq) ⇌ H+(aq) + A−(aq), which of these statements is true? For the generic equilibrium , which of
earnstyle [38]

Answer:

The correct answer is The equilibrium constant for this reaction changes as the pH changes.

Explanation:

The equilibrium constant of a reaction depends of pH and viceverca.There is a equation that link pH with equilibrium constant Ka(for assumption).The equation is given below,

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6 0
3 years ago
15 points
castortr0y [4]

Answer:

The number of proton of magnisium = 12

4 0
3 years ago
Recovery standards are a necessary tool for determining exactly how much of a particular analyte you are able to extract from a
Dafna11 [192]

Answer:

2.05mg Fe/ g sample

Explanation:

In all chemical extractions you lose analyte. Recovery standards are a way to know how many analyte you lose.

In the problem you recover 3.5mg Fe / 1.0101g sample: <em>3.465mg Fe / g sample. </em>As real concentration of the standard is 4.0 mg / g of sample the percent of recovery extraction is:

3.465 / 4×100 = <em>86,6%</em>

As the recovery of your sample was 1.7mg Fe / 0.9582g, the Iron present in your sample is:

1.7mg Fe / 0.9582g sample× (100/86.6) = <em>2.05mg Fe / g sample</em>

<em></em>

I hope it helps!

5 0
4 years ago
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