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xenn [34]
3 years ago
11

Help now please this is not to hard of a q

Chemistry
1 answer:
marishachu [46]3 years ago
3 0

Answer: Thomson

Explanation: Thomson proposed the plum pudding model of the atom.

You might be interested in
A 9.0 mL sample of oxygen gas (Q) is stored at a pressure of 6.3 atm. If the pressure of the sample
Julli [10]

Answer:

<h2>The answer is 23.63 mL</h2>

Explanation:

In order to find the new volume we use the formula for Boyle's law which is

P_1V_1 = P_2V_2 \\

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question

P1 = 6.3 atm

V1 = 9 mL

P2 = 2.4 atm

So we have

V_2 =  \frac{6.3 \times 9}{2.4}  \\  = 23.625

We have the final answer as

<h3>23.63 mL</h3>

Hope this helps you

6 0
3 years ago
Can i just have the answers
JulijaS [17]

https://www.teacherspayteachers.com/Product/Counting-Atoms-Worksheet-Editable-190185


id put em on here but i cant read them to good. Hope it helps tho!

fyi if u have other worksheets u need help on if u loo up the worksheet name and answer key itll prolly pop up.

3 0
3 years ago
Can someone please help me. Like right now please
EastWind [94]

Answer:

3rd law

Explanation:

when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

8 0
3 years ago
Calculate the number of moles in the 2.1-L volume of air in the lungs of the average person. Note that the air is at 37.0ºC (bod
adelina 88 [10]
Answer: 0.0826mol

PV=nRT
n=PV/RT
n=(1atm)(2.1L)/(310K)(0.082057L*atm/mol*K)=0.0826mol
7 0
3 years ago
Suppose a solution is prepared by dissolving 15.0 g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of O
Kipish [7]

Answer:

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

Explanation:

Moles of NaOH = \frac{15.0 g}{40 g/mol}=0.375 mol

Molarity of the nitric acid solution = 0.250 M

Volume of the nitric solution = 0.150 L

Moles of nitric acid = n

Molarity=\frac{Moles}{Volume(L)}

n=0.250 M\times 0.150 L=0.0375 mol

NaOH+HNO_3\rightarrow NaNO_3+H_2O

According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :

\frac{1}{1}\times 0.0375 mol of NaOH

Moles of NaOH left unreacted  in the solution =

= 0.375 mol - 0.0375 mol = 0.3375 mol

NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)

1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.

Then 0.3375 moles of NaOH will give :

1\times 0.3375 moles=0.3375 mol of hydroxide ion

The molarity of hydroxide ion in solution ;

=\frac{0.3375 mol}{0.150 L}=2.25 M

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

3 0
3 years ago
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