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3 years ago

Conductors allows electricity to flow easily through them. (examples copper and

Chemistry

2 answers:

poizon [28]3 years ago

7 0

The answer would be true almost all metals conduct electricity

Ivan3 years ago

5 0

Answer:

I believe its True

Explanation:

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1. Which of the following facts did Dalton realize?

Ne4ueva [31]

He discovered choice C! this is because atoms of the same element can have different numbers of neutrons

8 0

2 years ago

The largest pieces being dragged along the streambed are undergoing the process called ________. pushing saltation traction susp

lianna [129]

Answer:

saltation

Explanation:

Saltation is the transportation of particles by fluids such as by winds and water. It takes place when loose matter is removed from the sea bed and gets carried by the fluid back to the surface.

Examples include pebbles, sand, and soil. Saltation layers can form during avalanches.

8 0

3 years ago

What contribution did john dalton make to chemistry?

damaskus [11]

The son of a weaver, Dalton's major contribution to the field of chemistry is his atomic theory proposed in 1803.

5 0

3 years ago

When 2.00 g of methane are burned in a bomb calorimeter, the change in temperature is 3.08°C. The heat capacity of the calorimet

melisa1 [442]

Answer:

The approximate molar enthalpy of combustion of this substance is -66 kJ/mole.

Explanation:

First we have to calculate the heat gained by the calorimeter.

$q=c\times \Delta T$

where,

q = Heat gained = ?

c = Specific heat = $2.68 kJ/^oC$

ΔT = The change in temperature = 3.08°C

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

$q=2.68 kJ/^oC\times 3.08^oC$

$q=8.2544 kJ$

Now we have to calculate molar enthalpy of combustion of this substance :

$\Delta H_{comb}=-\frac{q}{n}$

where,

$\Delta H_{comb}$ = enthalpy change = ?

q = heat gained = 8.2544kJ

n = number of moles methane = $\frac{\text{Mass of methane}}{\text{Molar mass of methane }}=\frac{2.00 g}{16.042 g/mol}=0.1247 mole$

$\Delta H_{comb}=-\frac{8.2544 kJ}{0.1247 mole}=-66.21 kJ/mole\approx -66 kJ/mole$

Therefore, the approximate molar enthalpy of combustion of this substance is -66 kJ/mole.

6 0

3 years ago

Air should be classified as:

Alex777 [14]

Answer:

element

Explanation:

it is use full for you thanks

6 0

3 years ago