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3 years ago

A household variety of coal is

Chemistry

1 answer:

Lisa [10]3 years ago

3 0

Answer:

Anthracite coal

Explanation:

Anthracite coal has highest amount of carbon and has clean burning properties. So, it is used as a household coal.

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_____ waves are the p and s waves of an earthquake that shake the rocks beneath the surface as they travel along

MrRa [10]

Transverse? i think or longitudinal

5 0

3 years ago

10. A small gold nugget has volume of 0.87 cm3. What is its mass if the density of gold is 19.3 g/cm3?

bulgar [2K]

Answer:

16.791 grams

Explanation:

The density formula is:

$d=\frac{m}{v}$

Rearrange the formula for m, the mass. Multiply both sides of the equation by v.

$d*v=\frac{m}{v}*v$

$d*v=m$

The mass of the gold nugget can be found by multiplying the density and volume. The density is 19.3 grams per cubic centimeter and the volume is 0.87 cubic centimeters.

$d= 19.3 g/cm^3\\v-0.87 cm^3$

Substitute the values into the formula.

$m=d*v$

$m= 19.3 g/cm^3*0.87 cm^3$

Multiply. Note that the cubic centimeters, or cm³ will cancel each other out.

$m=19.3 g*0.87$

$m=16.791 g$

The mass of the gold nugget is 16.791 grams.

8 0

3 years ago

Atoms and elements are examples of A. molecules. B. mixtures. C. compounds. D. pure substances.

UkoKoshka [18]

Answer:

atom are very tiny partical

3 0

3 years ago

The rate constant for a certain reaction is k = 5.40×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the

IceJOKER [234]

Answer : The concentration after 17.0 minutes will be, $4.05\times 10^{-4}M$

Explanation :

The expression for first order reaction is:

$[C_t]=[C_o]e^{-kt}$

where,

$[C_t]$ = concentration at time 't' (final) = ?

$[C_o]$ = concentration at time '0' (initial) = 0.100 M

k = rate constant = $5.40\times 10^{-3}s^{-1}$

t = time = 17.0 min = 1020 s (1 min = 60 s)

Now put all the given values in the above expression, we get:

$[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}$

$[C_t]=4.05\times 10^{-4}M$

Thus, the concentration after 17.0 minutes will be, $4.05\times 10^{-4}M$

4 0

3 years ago

At a certain temperature the vapor pressure of pure heptane C7H16 is measured to be 454.mmHg. Suppose a solution is prepared by

OlgaM077 [116]

Answer:

Mass of heptane = 102g

Vapor pressure of heptane = 454mmHg

Molar mass of heptane = 100.21

No of mole of heptane = mass/molar mass = 102/100.21

No of mole of heptane = 1.0179

Therefore the partial pressure of heptane = no of mole heptane *Vapor pressure of heptane

Partial pressure of heptane = 1.0179*454mmHg

Partial pressure of heptane = 462.1096 = 462mmHg

the partial pressure of heptane vapor above this solution = 462mmHg

5 0

3 years ago