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3 years ago

Can anyone help me with the last question. it whould help me a ton

Chemistry

1 answer:

lina2011 [118]3 years ago

8 0

For the very last question you would first divide 8,900 by 12 and the number you get will be the answer. You should get 741.66

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For many purposes we can treat ammonia NH3 as an ideal gas at temperatures above its boiling point of −33.°C. Suppose the temper

Keith_Richards [23]

Answer:

The new pressure will be 0.225 kPa.

Explanation:

Applying combined gas law:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1\text{ and }V_1$ are initial pressure and volume at initial temperature $T_1$ .

$P_2\text{ and }V_2$ are final pressure and volume at initial temperature $T_2$ .

We are given:

$P_1=0.29 kPa\\V_1=V\\P_2=?\\V_2=V+50\% of V=1.5 V$

$T_1=-25^oC=248.15 K$

$T _2 = 16^oC=289.15 K$

Putting values in above equation, we get:

$\frac{0.29 kPa\times V}{248.15 K}=\frac{P_2\times 1.5V}{289.15 K}$

$P_1=0.225 kPa$

Hence, the new pressure will be 0.225 kPa.

4 0

3 years ago

Giving brainiest to whoever answers correctly.

Digiron [165]

Answer:250

Explanation:

5 0

3 years ago

When an electron is displaced in a semiconductor, the hole that's left behind is

geniusboy [140]

answer is A : attracted to the negative terminal of the voltage source
I think, that the "hole"moves as it captures a free electron leaving another hole in a slightly different place. The electron leaving leaves a net + charge, which is attracted to the negative terminal. Because the "hole" behaves as a positive charge it is attracted towards the negative terminal.

5 0

3 years ago

Read 2 more answers

How many atoms of N are in 0.82 g of NaNO3? ( molar conversion)

VMariaS [17]

The number of atoms N = 5.8 x 10²¹

<h3>Further explanation</h3>

A mole is a unit of many particles (atoms, molecules, ions) where 1 mole is the number of particles contained in a substance that is the same amount as many atoms in 12 gr C-12

1 mole = 6.02.10²³ particles

mass of N in 0.82 g of NaNO₃ (MW NaNO₃: 85 g/mol) :

$\tt mass~N=\dfrac{Ar~N}{MW~NaNO_3}\times mass~NaNO_3\\\\mass~N=\dfrac{14}{85}\times 0.82\\\\mass~N=0.135~g$