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3 years ago

13

In discharge tube gases are good conductors of electricity at very low pressure and high

1 answer:

Yuki888 [10]3 years ago

3 0

4)presence of ions and electrons

1)presence of atoms

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What fraction of atoms in morphine is accounted for by carbon?

Yuki888 [10]

To determine the fraction of carbon in morphine, we need to know the chemical formula of morphine. From my readings, the chemical formula would be <span>C17H19NO<span>3. We assume we have 1 g of this substance. Using the molar mass, we can calculate for the moles of morphine. Then, from the formula we relate the amount of carbon in every mole of morphine. Lastly, we multiply the molar mass of carbon to obtain the mass of carbon. We calculate as follows:

1 g </span></span> <span>C17H19NO<span>3 ( 1 mol / 285.34 g ) ( 17 mol C / 1 mol </span></span> <span>C17H19NO3</span>) ( 12.01 g C / 1 mol C) = 0.7155 g C

Fraction of carbon = 0.7155 g C / 1 g <span>C17H19NO<span>3 = 0.7155</span></span>

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3 years ago

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This equation shows the combustion of methanol.

sveticcg [70]

Answer: 44.8 L

Explination:

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3 years ago

What is the best definition of energy?

tatuchka [14]

Answer:

C

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5 0

3 years ago

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Plz help me with this question plzzzzz

LekaFEV [45]

Answer:

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Explanation:

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3 years ago

A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure

Lesechka [4]

Answer:

the final volume of the gas is $V_2$ = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure $P_{ext}$ :

$P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}$

$P_{ext} = 1.03 \ atm$

The workdone W = $P_{ext}$ V

The change in volume ΔV= $\dfrac{W}{P_{ext}}$

ΔV = $\dfrac{130.1 \ J \times \dfrac{1 \ L \ atm}{ 101.325 \ J} }{1.03 \ atm }$

ΔV = $\dfrac{1.28398717 }{1.03 }$

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial volume = 61.5 mL

The change in volume V is $\Delta V = V_2 -V_1$

$- V_2= - \Delta V -V_1$

multiply through by (-), we have:

$V_2= \Delta V+V_1$

$V_2$ = 1250 mL + 61.5 mL

$V_2$ = 1311.5 mL

∴ the final volume of the gas is $V_2$ = 1311.5 mL

5 0

3 years ago