Im pretty sure it weight because when we are on different plants with a different amount of gravitational pull our weight changes
hope this helps :)
Answer:
32(molecular mass has no unit )
Explanation:
(16)(o2)
16×2
=32
Answer:
The molarity of the solution is 1,03 M.
Explanation:
Molarity is a concentration measure that expresses the moles of solute (in this case HBR) in 1 liter of solution (1000ml). First we calculate the mass of 1 mol of HBr, to calculate the moles that are in 50 g of said compound:
Weight 1 mol HBr= Weight H + Weight Br= 1,01g + 79,90g= 80, 91 g/mol
80,91 g ----1 mol HBr
50,0 g------x= (50,0 g x1 mol HBr)/80,91 g= 0,62 mol HBr
600 ml solution-----0,62 mol HBr
1000ml solution------x= (1000ml solution x 0,62 mol HBr)/600 ml solution
<em>x=1,03 moles HBr ---> The solution is 1,03M</em>
Answer:
V KOH = 41 mL
Explanation:
for neutralization:
- ( V×<em>C </em>)acid = ( V×<em>C </em>)base
∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L
∴ V H2SO4 = 41 mL = 0.041 L
∴ <em>C</em> KOH = 0.0050 N = 0.0050 eq-g/L
∴ E KOH = 1 eq-g/mol
⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L
⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH
⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)
⇒ V KOH = 0.041 L
Answer:
is the formula for the limiting reagent.
Mass of silver chloride produced is 71.8 g.
Explanation:
Moles of silver nitrate = 0.500 mol
Moles of copper(II) chloride = 0.285 mol
According to reaction, 2 moles of silver nitrate reacts with 1 mole of copper chloride , then 0.500 mole of silver nitrate will react with :
of copper(II) chloride
As we can see that moles of copper(II) chloride will be reacting is 0.250 mol less than present moles of copper (II) chloride ,so this means that silver nitrate is limiting reagent.
And moles of silver chloride to be formed will depend upon silver nitrate.
According to reaction, 2 moles of silver nitrate gives 2 moles of silver chloride , then 0.500 mole of silver nitrate will give :
of silver chloride
Mass of silver chloride produced:
0.500 mol × 143.5 g/mol = 71.8 g
