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irga5000 [103]
3 years ago
7

Most atoms normally have a net charge that is _________.

Chemistry
1 answer:
Citrus2011 [14]3 years ago
7 0
Zero because there are so many electrons as protons.
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What is the mass in grams of 10 moles of ammonia, NH3?
8_murik_8 [283]

Answer:

C. 170 g.

Explanation:

multiply given moles by the molar mass of ammonia.

6 0
2 years ago
Which of these are neutral salts? Check all that apply.
Elden [556K]
The following are neutral salts: NANO3, BaBr2 AND Ba[OH]2. 
Neutral salts are those salts that are formed when a strong base react with a strong acid. In those neutralization reactions, the base completely neutralize the acid to form a neutral salts.
8 0
3 years ago
Read 2 more answers
2. Which types of changes observe the law of
mart [117]
D) physical and chemical changes
3 0
3 years ago
Select the correct answer from each drop-down menu.
Alika [10]

Answer:

Explanation:

There are three fundamentally known states of matter save for plasma and the Bose-Einstein condensate. These states of matter are solid, liquid and gas.

We can classify nearly all matter into these discrete categories based on certain lucid characteristics they exhibit.

  • Solids have definite shape and volume and they exhibit by the far the most remarkable internal ordering. Their molecules are attached by strong internal forces.
  • Liquids do not have a definite shape but takes the shape of the contains that hold them. They are not random and have internal cohesion among them.
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4 0
3 years ago
Starting from a 1 M stock of NaCl, how will you make 50 ml of 0.15 M NaCl?
WARRIOR [948]

Answer:

We need 7.5 mL of the 1M stock of NaCl

Explanation:

Data given:

Stock = 1M this means 1 mol/ L

A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL

Step 2: Calculate the volume of stock we need

The moles of solute will be constant

and n = M*V  

M1*V1 = M2*V2

⇒ with M1 = the initial molair concentration = 1M

⇒ with V1 = the volume we need of the stock

⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L

⇒ with M2 = the concentration of the new solution = 0.15 M

1*V1 = 0.15*(50)

V1 = 7.5 mL

Since 0.0075 L of 1M solution contains 0.0075 moles

50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M

We need 7.5 mL of the 1M stock of NaCl

6 0
2 years ago
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