<h2>
1.25 g of
would be produced from the complete reaction of 25 mL of 0.833 mol/L
with excess
</h2>
Explanation:
To calculate the number of moles for given molarity, we use the equation:
![0.833M=\frac{\text{Moles of} HC_3H_3O_2\times 1000}{25ml}\\\\\text{Moles of} HC_3H_3O_2 =\frac{0.833mol/L\times 25}{1000}=0.0208mol](https://tex.z-dn.net/?f=0.833M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%7D%20HC_3H_3O_2%5Ctimes%201000%7D%7B25ml%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%7D%20HC_3H_3O_2%20%3D%5Cfrac%7B0.833mol%2FL%5Ctimes%2025%7D%7B1000%7D%3D0.0208mol)
![NaHCO_3+HC_2H_3O_2\rightarrow NaC_2H_3O_2+H_2O+CO_2](https://tex.z-dn.net/?f=NaHCO_3%2BHC_2H_3O_2%5Crightarrow%20NaC_2H_3O_2%2BH_2O%2BCO_2)
According to stoichiometry:
1 mole of
will give = 1 mole of ![CO_2](https://tex.z-dn.net/?f=CO_2)
0.0208 moles of
will give =
of ![CO_2](https://tex.z-dn.net/?f=CO_2)
Mass of ![HC_2H_3O_2=moles\times {\text {molar mass}}=0.0208\times 60g/mol=1.25g](https://tex.z-dn.net/?f=HC_2H_3O_2%3Dmoles%5Ctimes%20%7B%5Ctext%20%7Bmolar%20mass%7D%7D%3D0.0208%5Ctimes%2060g%2Fmol%3D1.25g)
Thus 1.25 g of
would be produced from the complete reaction of 25 mL of 0.833 mol/L
with excess
Learn more about molarity
https://brainly.in/question/13034158
#learnwithbrainly
Answer:
being responsible is owing up to the possible
consequences of your decision whether it was right or wrong.
communication is a two way process in which there is a messenger and receiver
one of the ways by which a person could be responsible is by saying 'I'
in all the things he would say, instead of 'we'
for example , " We don't like the way you acted a while ago " , change it to i don't like tye way you acted a while ago.
here u r establishing ownership on the things you want to say to the person
From the equation:
4mol Li react with 1 mol O2
Molar mass Li = 7g/mol
mol in 84g Li = 84/7 = 12 mol Li
From the equation - 12 mol Li will react with 3 mol O2
At STP 1 mol O2 has volume = 22.4L
<span>
At STP 3 mol O2 has volume = 3*22.4 = 67.2L O2 gas will react. </span>
Answer:
1 : 1.5
Explanation:
First Sample;
Ratio of sulfur and Oxygen
Mass of sulfur : Mass of oxygen
Mass of oxygen = Mass of sample - Mass of sulfur = 70 - 35 = 35g
35g : 35g
1 : 1
Second Sample;
Ratio of sulfur and Oxygen
Mass of sulfur : Mass of oxygen
Mass of oxygen = Mass of sample - Mass of sulfur = 70 - 28 = 42g
28g : 42g
1 : 1.5
Further reducing it to make oxygen 1;
0.6667 : 1
ratio in whole numbers of the masses of sulfur that combine with 1.00 g of oxygen between the two compounds;
0.6667 : 1
1 : 1.5
Answer:
6.022×1023
Here, it has been asked how many eggs are in one mole of eggs. The eggs represent the unit which can be atom, ion or molecule. So by the definition of mole, 6.022×1023number of eggs are present in one mole of eggs.