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IRISSAK [1]
3 years ago
7

A system of gas at low density has an initial pressure of 1.90 × 10 5 1.90×105 Pa and occupies a volume of 0.18 m³. The slow add

ition of 965 J of heat to the system causes it to expand isobarically to a volume of 0.51 m³. What is the change in the internal energy of the system?
Chemistry
1 answer:
Bad White [126]3 years ago
8 0

Answer:

ΔU = -6.2 × 10⁴ J

Explanation:

The system absorbs 965 J of heat, that is, q = 965 J.

The work (w) can be calculated using the following expression.

w = -P . ΔV

where,

P is the external pressure

ΔV is the change in the volume

w = - (1.90 × 10⁵ N/m²) × (0.51 m³ - 0.18 m³) = -6.3 × 10⁴ J

The change in the internal energy (ΔU) is:

ΔU = q + w = 965 J + (-6.3 × 10⁴ J) = -6.2 × 10⁴ J

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4 0
3 years ago
Eli states that sodium phosphate is a mixture because it is composed of both sodium ions and phosphate ions. Which is the best a
Nikolay [14]

Answer:

The question is incomplete because the options are not given, here are the options gotten from another websites.

a. It is correct because each ion is a pure substance, so sodium phosphate is made up of two pure substances.

b. It is correct because the composition of sodium phosphate changes depending on the sample.

c. It is incorrect because sodium phosphate is a compound that has a single composition.

d. It is incorrect because the two types of ions in sodium phosphate cannot be seen.

The correct option is C.

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Explanation:

It is incorrect because sodium phosphate is a compound that has a single composition because a Compound is substance that contain two or elements which combine together by chemical bonds or the atoms of the element are joined together by chemical bonds. From our question, sodium phosphate is a compound because it is composed of two elements which are sodium and phosphate and these elements atoms chemically combine together.

5 0
3 years ago
Read 2 more answers
What is the pOH of 2.5 M NaOH
victus00 [196]

Answer:

-0.398

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pOH=-\log \left [ OH^- \right ]

To find pOH of 2.5 M NaOH, solve pOH=-\log \left [ 2.5 \right ] = -0.398

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3 years ago
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