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Brilliant_brown [7]
3 years ago
6

What is a cell GJFFVf xfhdfvcdd

Chemistry
2 answers:
andreev551 [17]3 years ago
6 0

Answer:

Yeah

Explanation:

balu736 [363]3 years ago
5 0
In biology, the smallest unit that can live on its own and that makes up all living organisms and the tissues of the body’s a cell. A cell has three main parts: the cell membrane, the nucleus, and the cytoplasm. ... The nucleus is a structure inside the cell that contains the nucleolus and most of the cell's DNA.

Hope that helps
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How many moles of carbon are in 3.5 moles of calcium carbonate? ______?
DerKrebs [107]
<span>3.5 moles CaCO3 (1 mole carbon/1 mole CaCO3) = 3.5 moles</span>
8 0
3 years ago
Read 2 more answers
What is the atomicity of ozone
faltersainse [42]

Answer:

Explanation:

Atomicity of an element is a measure of the total number of atoms present in a molecule. For example- O2 , O Similarly, an ozone molecule consists of 3 atoms of oxygen and has an atomicity of 3. The atomicity of ammonia is NH3.Ammonia is a compound of nitrogen and hydrogen with the formula NH3.

(Could very much be incorrect, it's just what I think it is.)

8 0
4 years ago
Calculate the mass (in grams) of methylene bluecrystals that you must weigh in order to make 100.0mL of 1.25 × 10-5mol/L methyle
mario62 [17]

<u>Answer:</u> The mass of methylene blue that must be weighed is 3.99\times 10^{-4}g

<u>Explanation:</u>

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = 1.25\times 10^{-5}M

Molar mass of methylene blue = 319.85 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

1.25\times 10^{-5}M=\frac{\text{Mass of methylene blue}\times 1000}{319.85\times 100.0}\\\\\text{Mass of methylene blue}=\frac{1.25\times 10^{-5}\times 319.85\times 100.0}{1000}=3.99\times 10^{-4}g

Hence, the mass of methylene blue that must be weighed is 3.99\times 10^{-4}g

4 0
3 years ago
Use the following equation to answer the questions and please show all work.
DIA [1.3K]

Answer:

a.36 g of water is produced.

b.64 g of O_{2} is consumed.

Explanation:

The reaction is 2H_{2} + O_{2}⇒2H_{2}O

a.

Given,

Weight of H_{2} reacted = 4g

Weight of 1 mole of H_{2} = 2\times1 = 2g

Therefore no. of moles of H_{2} reacted = \frac{4}{2} = 2 moles;

Also given,

Weight of O_{2} reacted = 32 g

Weight of 1 mole of O_{2} = 2\times16 = 32 g

Therefore no. of moles of O_{2} reacted = \frac{32}{32} = 1

We know that 2 moles of Hydrogen reacts with 1 mole of Oxygen to give 2 moles of water,

As we took 2 moles of Hydrogen and 1 mole of Oxygen,

Directly,from the equation we can tell 2moles of water will be produced.

Therefore no. of moles of H_{2} O produced = 2

Weight of 1 mole of water = 2\times 1+16 = 18

Therefore weight of H_{2}O produced = 2\times 18 = 36gm

b.

Given ,

72 g of H_{2}O is produced.

So,

no. of moles of H_{2}O produced =\frac{72}{18} = 4 moles

From equation For every 2 moles of water formed , 1 mole of oxygen must be required.

So for producing 4 moles of water,

No. of moles of Oxygen required = 2 moles.

Therefore weight of O_{2} reacted = 2\times32 = 64 g

Method 2:

Given,

8 g of H_{2} has reacted.

So,

no. of moles of H_{2} reacted = \frac{8}{2} = 4 moles.

From equation , we know that For every 2 moles of H_{2} reacted,1 mole of O_{2} will react.

Therefore,

No. of moles of O_{2} that reacts with 4 moles of H_{2} = 2\times1 = 2 moles

Therefore the weight of O_{2} reacted = 2\times 32 = 64 g

6 0
3 years ago
When the hydronium ion concentration equals 10 moles per liter, what is the ph of the solution? is the solution acidic or basic?
Nata [24]

pH = - log[acid] = -log [H+]

pH= -log[10] = 1

Thus, pH is 1

if pH is 7, the solution is neutral and if pH is greater than 7 the solution is basic and if it is less than 7 the solution is acidic.

This solution have pH = 1, thus the solution is acidic.


4 0
3 years ago
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