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Harman [31]
3 years ago
9

Hydrogen has three isotopes 1H, 2H, and 3H. What is the difference between these three isotopes?

Chemistry
1 answer:
Leona [35]3 years ago
4 0
The main difference between the 3 isotopes of hydrogen are the number of neutrons in the nucleus. Hydrogen has no neutrons, Deuterium has one neutron, and tritium has two neutrons. All three have one proton and one electron.
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What is molarity of a solution containing 0.34 moles of FeCl3 in the 0.45 liters of water
Minchanka [31]
Hello!

Molarity is defined as the number of moles of a substance that can be found in 1 liter of solution. That is expressed mathematically as M=mol/L

For the FeCl₃ solution:

M= \frac{mol FeCl_3}{L sol}= \frac{0,34 moles FeCl_3}{0,45 L}= 0,756 M

So, the concentration of this solution is 0,756 M

Have a nice day!
7 0
3 years ago
A compound is made up of at least
solmaris [256]
Two or more different elements
5 0
3 years ago
Number the elements sodium, magnesium, and potassium in the predicted order of ionic radius from the largest (1) to the smallest
Aleksandr-060686 [28]
Potassium ,sodium, and magnesium
7 0
3 years ago
Read 2 more answers
onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0
3 years ago
An analysis of a sugar compound is found to contain 132 grams of carbon (C), 22 grams of hydrogen (H), and 176 grams of oxygen (
ddd [48]
132 g of C  ,   22 g of H   , 176 g of O

132 + 22 + 176 => 330 g <span>of the substance

</span>Now convert the masses in <span>moles :
</span>
C = 12.0 u        H = 1.0 u       O = 16.0 u

C = 132 / 12.0 => 11 moles

H = 22 / 1.0 => 22 moles

O = 176 / 16.0 => 11 moles

Using the values obtained the lowest proportion in mols of elements present, simply divide the values found for the least of them<span>:
</span>
C = 11 / 11 => 1

H = 22 / 11 => 2

O = 11 / 11 => 1

 formula empirically <span>is : CH</span>₂O

hope this helps!


8 0
3 years ago
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