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salantis [7]
3 years ago
13

Can someone plz help me with this one problem plzzzzz I’m being timed!!!

Chemistry
1 answer:
poizon [28]3 years ago
4 0

Answer:

answer : the penguins do

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You have 500 mL of 5 M HCl already made. You need to dilute the solution to 1 M HCl. How much water will you need to add?
hram777 [196]

Answer:

You need to add 400mL of water

Explanation:

500mL = 5 M HCI     That means that if you divide both sides by 5

100mL = 1 M HCI         If you need ot get rid of 4 M HCI then you add 400 mL of water because that is what it is equal to

5 0
3 years ago
How many moles are in 5.03 x 1023 molecules?
AlladinOne [14]

Answer:

Some formulas for calculating mole are

Mole = Mass/ Molar mass

Mole = no of particles / avogadros constant

NB : no of particles can be no of atoms , no of ions , or no of molecules 2. Avogadros number or constant = 6.02 times 10 ^23

so we will be using the second formula

Mole = no of particles / avogadros constant

Mole =  5.03 x 10 ^23/6.02 x10^23

Mole = 8.355x10^45

hope it helps :)

Explanation:

4 0
3 years ago
Help plssss <br><br> The question is in the picture above
Paul [167]

Answer:

The answer is : A

8 0
3 years ago
Read 2 more answers
Objects 1 and 2 collide and stick together. Which best describes the momentum of the resulting single object? It is less than th
Mariulka [41]

Answer:

The answer would be D

Explanation:

Hope I helped <3

5 0
3 years ago
Read 2 more answers
What is the mass in grams of 5.00moles of CH4?
anastassius [24]

Answer:

1. 80g

2. 1.188mole

Explanation:

1. We'll begin by obtaining the molar mass of CH4. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Number of mole of CH4 from the question = 5 moles

Mass of CH4 =?

Mass = number of mole x molar Mass

Mass of CH4 = 5 x 16

Mass of CH4 = 80g

2. Mass of O2 from the question = 38g

Molar Mass of O2 = 16x2 = 32g/mol

Number of mole O2 =?

Number of mole = Mass /Molar Mass

Number of mole of O2 = 38/32

Number of mole of O2 = 1.188mole

6 0
3 years ago
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