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3 years ago

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

1 answer:

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 of the substance

Now convert the masses in moles :

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:

C = 11 / 11 => 1

H = 22 / 11 => 2

O = 11 / 11 => 1

formula empirically is : CH₂O

hope this helps!

8 0

3 years ago