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2 years ago

The formula for water is H20. What does the formula H2O tell us about the composition of water?

1 answer:

4 0

H2O is the chemical formula of water. It means that each molecule of water is made up of two hydrogen atoms, indicated by the letter H, and a single oxygen atom, represented by the O. Water is a chemical substance with no smell, taste, or color.

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What happens in a stationary front?

lions [1.4K]

Answer:

c a cold air mass and a warm air mass meet with neither moving

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3 years ago

What is the final volume of NaOH solution prepared from 100.0 mL of 0.500 M NaOH if you wanted the final concentration to be 0.1

suter [353]

Answer:

The final volume of NaOH solution is 30ml

Explanation:

We all know that

V1S1 = V2S2

or V1= V2S2÷S1

or V1= V2×S2×1/S1

or V1=100×0.15×1/0.50

V1= 30

∴30 ml NaOH solution is required to prepare 0.15 M from 100ml 0.50 M NaOH solution.

5 0

3 years ago

Which phrases describe sedimentary rock? check all that apply.

yuradex [85]

Answer:

Sedimentary rock, rock formed at or near Earth's surface by the accumulation and lithification of sediment by the precipitation from solution at normal surface temperatures

Explanation:

7 0

2 years ago

The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

hammer [34]

Answer:

A) 1059 J/mol

B) 17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

$C_p=C_v+R$

We can determine $C_v$ from above if we make $C_v$ the subject of the formula as:

$C_v$ $=C_p-R$

$C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314$

$C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4$

A).

The formula for calculating change in internal energy is given as:

$dU=C_vdT$

If we integrate above data into the equation; it implies that:

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 1059J/mol$

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 17,920 J/mol$

3 0

3 years ago

What is the study of clouds called?

Kamila [148]

The answer is Nephology

5 0

3 years ago