1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
seropon [69]
3 years ago
13

An aqueous salt solution is 15.0% mass sodium chloride. How many grams of salt are in 250.0 grams of this solution? Use correct

Chemistry
1 answer:
Shtirlitz [24]3 years ago
5 0

Answer:

37.5 g NaCl

Explanation:

Step 1: Given data

  • Concentration of NaCl: 15.0% m/m
  • Mass of the solution: 250.0 g

Step 2: Calculate how many grams of NaCl are in 250.0 g of solution

The concentration of NaCl is 15.0% by mass, that is, there are 15.0 g of NaCl every 100 g of solution.

250.0 g Solution × 15.0 g NaCl/100 g Solution = 37.5 g NaCl

You might be interested in
Which of the following values are not equal to 1 mole?
Gekata [30.6K]

Answer:

none of them are equal to one mole

3 0
3 years ago
What changes chemical energy to something you can use
lutik1710 [3]

based on the law of conservation of energy its it atoms hold by strong chemical bonds

6 0
3 years ago
What does a nubula turn into
LekaFEV [45]

Answer:

a white dwarf

Explanation:

6 0
3 years ago
When an atom of calcium (Ca) forms an ion by losing two electrons, what is the ion’s type and charge?
Svetllana [295]

Answer: it would be cation, 2+

Explanation: electrons are negatively charged by 1. So if you get rid of 2 electrons it would be positive and cation is used to represent positive ions.

8 0
3 years ago
Read 2 more answers
How many grams N2F4 can be produced from 225 g F,?​
zavuch27 [327]

Answer:

308 g

Explanation:

Data given:

mass of Fluorine (F₂) = 225 g

amount of N₂F₄ = ?

Solution:

First we look to the reaction in which Fluorine react with Nitrogen and make N₂F₄

Reaction:

          2F₂ + N₂ -----------> N₂F₄

Now look at the reaction for mole ratio

          2F₂     +    N₂   ----------->  N₂F₄

        2 mole                              1 mole

So it is 2:1 mole ratio of Fluorine to N₂F₄

As we Know

molar mass of F₂ = 2(19) = 38 g/mol

molar mass of N₂F₄ = 2(14) + 4(19) =

molar mass of N₂F₄ = 28 + 76 =104 g/mol

Now convert moles to gram

                 2F₂          +       N₂   ----------->  N₂F₄

        2 mole (38 g/mol)                        1 mole (104 g/mol)

                 76 g                                           104 g

So,

we come to know that 76 g of fluorine gives 104 g of N₂F₄ then how many grams of N₂F₄ will be produce by 225 grams of fluorine.

Apply unity formula

                  76 g of F₂ ≅ 104 g of N₂F₄

                   225 g of F₂ ≅ X of N₂F₄

Do cross multiplication

                  X of N₂F₄ = 104 g x 225 g / 76 g

                  X of N₂F₄ = 308 g

So,

308 g N₂F₄ can be produced from 225 g F₂

7 0
3 years ago
Other questions:
  • Which element can expand its valence shell to accommodate more than eight electrons? which element can expand its valence shell
    14·2 answers
  • Which of the following is a characteristic property of noble gases? A. They only react with each other B. They do not react chem
    8·1 answer
  • after a chemistry student made a AgNO3(small 3 at bottom) solution., she wanted to determine the molar concentration of it. if 2
    8·1 answer
  • I beg you please please I beg you help me please I’m litterly crying please :(
    6·1 answer
  • Four processes that change rock from one type to another.
    6·1 answer
  • One of the hydrates of MnSO4 is manganese(II) sulfate tetrahydrate . A 71.6 gram sample of MnSO4 4 H2O was heated thoroughly in
    7·1 answer
  • How many moles are in 3.85*10^4 grams of Ba(NO3)2
    10·2 answers
  • Helppppppppppppppppp
    5·1 answer
  • In covalent bonds atmosphere share electrons so that?
    9·1 answer
  • When potassium permanganate is stirred in water, it turns to purple. How would
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!