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Dominik [7]
3 years ago
7

How many joules are required to heat 120.5!grams of water from 5.6°C to 62.5°C

Chemistry
1 answer:
DENIUS [597]3 years ago
6 0

Answer:

The amount of energy required is 17 kJ.

Explanation:

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Air has a mass of 1.2g and a volume of 4,555ml. What is the density
coldgirl [10]

Answer:

<h2>Density = 0.00026 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3>Density(\rho) =  \frac{mass}{volume}</h3>

From the question

mass of air = 1.2 g

volume = 4,555 mL

Substitute the values into the above formula and solve for the density

That's

<h3>Density =  \frac{1.2}{4555}</h3>

= 0.0002634

We have the final answer as

<h3>Density = 0.00026 g/mL</h3>

Hope this helps you

3 0
3 years ago
How do you find the highest energy level?
joja [24]

Answer:

The larger the number of the energy level, the farther it is from the nucleus. Electrons that are in the highest energy level are called valence electrons. Within each energy level is a volume of space where specific electrons are likely to be located.

4 0
2 years ago
What are the evidence of continental drift
Veronika [31]
Widespread distribution of Permo-Carboniferous glacial sediments in South America, Africa, Madagascar, Arabia, India, Antarctica and Australia was one of the major pieces of evidence for the theory of continental drift.
5 0
3 years ago
Read 2 more answers
7. Balance the following reaction under basic conditions: Al (s) + Cr2O72- (aq) -&gt; Al3+ (aq) + Cr3+ (aq)
MArishka [77]

The free energy change of the reaction; Fe (s) + Au3+ (aq) -> Fe3+ (aq) + Au (s) is calculated to be -443.83KJ/mol.

For the reaction shown in question 7, we can divide it into half equations as follows;

Oxidation half equation;

6 Al (s) -------> 6Al^3+(aq) + 18e

Reduction half equation;

3Cr2O7^2-(aq) + 42H^+(aq)   + 18e -----> 6Cr^3+(aq) + 21H2O(l)

The balanced reaction equation is;

6Al(s) + 3Cr2O7^2-(aq) + 42H^+(aq)   -----> 6Al^3+(aq) + 6Cr^3+(aq) + 21H2O(l)

The E° of this reaction is obtained from;

E° anode = -0.04 V

E°cathode = +1.50 V

E° cell = +1.50 V - (-0.04 V) = 1.54 V

Given that;

ΔG° = -nFE°cell

n = 3, F = 96500, E°cell = 1.54 V

ΔG° = -(3 × 96500 × 1.54)

ΔG° = -443.83KJ/mol

Learn more: brainly.com/question/967776

4 0
3 years ago
What are some example of nuclear acids
vladimir1956 [14]
First of all, it is a nucleic acid, not nuclear acid.

Examples are DNA and RNA( Deoxyribonucleic Acid and Ribonucleic Acid)
3 0
3 years ago
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