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3 years ago

I give up on chem lol

Chemistry

2 answers:

elena55 [62]3 years ago

4 0

Same here this is to much :(

rusak2 [61]3 years ago

4 0

Same i gave up since a week ago

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When the number of protons and electrons are equal

Tomtit [17]

Answer:

when the atom is neutral

Explanation:

8 0

3 years ago

A sample of n2 occupies 3. 0l at 3. 0atm. What volume will it occupy when the pressure is changed to 0. 5atm and the temperature

belka [17]

It will occupy 18 L volume when the pressure is changed to 0. 5 atm and the temperature remain constant.

Calculation,

According to Boyle law, the pressure of a given quantity of gas varies inversely with its volume at constant temperature.

$P_{1} V _{1} = k$ .....(i)

$P_{1}$ is initial pressure = 3 atm
$V _{1}$ is initial volume = 3L
k is constant

$P_{2} V _{2} = k$ (ii)

$P_{2}$ is final pressure = ?
$V _{2}$ is final volume = 0.5atm
k is constant

Combining equation (i) and (ii). we get,

$P_{1} V _{1} = P_{2} V _{2}$

3 atm× 3L= $V _{2}$ × 0.5atm

$V _{2}$ = 3 atm× 3L/0.5atm = 18 L

Boyle law used during respiration or breathing.

To Learn more about Boyle law

brainly.com/question/1437490

#SPJ4

7 0

2 years ago

The relationship between a law and a theory is

irinina [24]

I would say it’s ‘laws are based on complex theories’.

A scientific law predicts the outcome, while a theory presents a possible explanation to why this outcome is. A law is based of a theory that supports the most data.

7 0

3 years ago

Answer these for me please

LUCKY_DIMON [66]

Answer:

13- a i think

Explanation:

5 0

4 years ago

Read 2 more answers

What is the volume of 3.2 moles of chlorine gas (Cl2) at 295 K and 1.1 atm?

Lostsunrise [7]

Answer:

$\boxed{\text{70 L}}$

Explanation:

We can use the Ideal Gas Law to solve this problem

pV = nRT

Data:

p = 1.1 atm

n = 3.2 mol

R = 0.082 16 L·atm·K⁻¹mol⁻¹

T = 295 K

Calculation:

$\begin{array}{rcl}1.1V\text{ atm} & = & \text{3.2 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{295 K}\\1.1V & = & \text{77.5 L}\\\\V & = &\dfrac {\text{77.5 L}}{1.1}\\\\V & = & \text{70 L}\\\end{array}\\\text{The volume is }\boxed{\textbf{70 L}}$

3 0

3 years ago