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tensa zangetsu [6.8K]
3 years ago
7

What is the name of the acid made from hydrogen and NO3 1-?

Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
7 0

Answer:

nitric acid  - HNO3

Explanation:

Hope it's correct! :)

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A liquid occupies a volume of 5.0 L has a mass of 6.0 kg. what is the density of a the lights in kg/L
grandymaker [24]

Answer:

Hey there! :

Volume = 5.0 L

mass = 6.0 Kg

Therefore:

Density = mass / Volume

Density = 6.0 / 5.0

Density = 1.2 kg/L

6 0
3 years ago
Can u help me with this
dolphi86 [110]

Answer:

c

Explanation:

the reason is because your a homosiejenenddj

8 0
2 years ago
Which organelle in the table is correctly matched with its function?<br> which one is it??
Papessa [141]

Answer:

\huge\boxed{\sf Ribosomes}

Explanation:

<h3>Organelles and their function:</h3><h3><u>Lysosomes:</u></h3>
  • Lysosomes functions in the digestion of food of the cell.
  • It contains hydrolytic enzymes.
<h3><u>Vacuole:</u></h3>
  • Vacuole mostly functions in storage.
<h3><u>Mitochondrion:</u></h3>
  • Mitochondrion is the power house of the cell.
<h3><u>Ribosome:</u></h3>
  • Ribosome functions in protein synthesis.

\rule[225]{225}{2}

4 0
2 years ago
What Happens when the temperature of an object increases?
Pepsi [2]
D) The objects particles move closer together.
6 0
3 years ago
Read 2 more answers
A 5.00 mL sample of hydrochloric acid is titrated with 0.1293 M ammonia (a base). If the titration required 28.15 mL of ammonia,
Sladkaya [172]

Answer:

1. C = 0.73 M.

2. pH = 0.14

       

Explanation:

The reaction is the following:

HCl + NH₃ ⇄ NH₄⁺Cl⁻

From the titration, we can find the number of moles of HCl that were neutralized by the ammonia.

n_{a} = n_{b}

Where "a" is for acid and "b" is for base.

The number of moles is:

n = C*V  

Where "C" is for concentration and "V" for volume.

C_{a}V_{a} = C_{b}V_{b}

C_{a} = \frac{0.1293 M*28.15 mL}{5.00 mL} = 0.73 M

Hence the initial concentration of the acid is 0.73 M.

The original pH of the acid is given by:

pH = -log([H^{+}])

pH = -log(0.73) = 0.14          

Therefore, the original pH of the acid is 0.14.

I hope it helps you!                  

4 0
3 years ago
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