How many grams of H20 is formed with 0.75 moles of NaOH

Answers

Answer 1
Answer:

Answer: 13.5g

Explanation:

Moles = mass/molar mass

Molar mass of water= 18gmol-1

0.75=m/18

m= 17*0.75= 13.5g


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A helium nucleus is formed from hydrogen nuclei. What nuclear change has occured? a.) beta decay b.) none c.) alpha decay d.) nuclear fusion​

Answers

Answer:

alpha decay

Hope this helps. :)

When salt dissolves in water, which is the solvent

Answers

a solvent is what is doing the dissolving. so in this case the water would be your solvent because it is dissolving the salt. does that make sense?

What are the two fundamental portions of an atom?​

Answers

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flight of a bald eagle is about 50 km a scientist measured an eagle flying 80 km how is this possible

Answers

The AVERAGE flight of a bald eagle is 50 km/h. so that means it can go higher or lower than that speed. it's not that it can only go a maximum of 50 km/h 
Hope this helps. :)

CAN ANYONE ANSWER THIS PLZZZ ?!?

Answers

Answer:you need to write the equation of the reactions in each question.

you need to calculate the mol for each substance which mass for which mass is given

then you do mol ratio.

I hope this shows you a hint and helps you not to depend completely on brainly.

plz put me as Brainliest

Explanation:

A compound contains 70.18 % oxygen by mass. What mass of this compound would be needed to obtain 48.45 grams of oxygen?

Answers

Answer:

                     69.036 g of Compound is required to obtain 48.45 g of Oxygen.

Explanation:

                    Let us assume that the total mass of the compound is 100 g. So it means that this compound weighing 100 is composed 70.18 % of Oxygen or among 100 g of this compound 70.18 g is constituted by oxygen only. Hence, we can make a relation as that,

                      70.18 g O is present in  =  100 g of a Compound

So,

         48.45 g of O will be present in  =  X g of a Compound

Solving for X,

                        X  =  100 g × 48.45 g ÷ 70.18 g

                        X  =  69.036 g of Compound