Answer:
separare pâlnii de picurare balon cotat. 1. 2. 3. 4 5 6 7. 8. 9. 1 ... Mod de lucru: 25 g Na2SO4∙7H2O se dizolvă în cantitatea minim ... Exemple: NaOH – hidroxid de sodiu.
Explanation:
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Answer:
the answer is D
Explanation:
percentage composition= mole of the substance divided by the total molar mass of the compound multiplied by 100.
Answer:
1.73 M
Explanation:
We must first obtain the concentration of the concentrated acid from the formula;
Co= 10pd/M
Where
Co= concentration of concentrated acid = (the unknown)
p= percentage concentration of concentrated acid= 37.3%
d= density of concentrated acid = 1.19 g/ml
M= Molar mass of the anhydrous acid
Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1
Substituting values;
Co= 10 × 37.3 × 1.19/36.5
Co= 443.87/36.6
Co= 12.16 M
We can now use the dilution formula
CoVo= CdVd
Where;
Co= concentration of concentrated acid= 12.16 M
Vo= volume of concentrated acid = 35.5 ml
Cd= concentration of dilute acid =(the unknown)
Vd= volume of dilute acid = 250ml
Substituting values and making Cd the subject of the formula;
Cd= CoVo/Vd
Cd= 12.16 × 35.5/250
Cd= 1.73 M
It would be the needle pressure gage temp check
Answer:
The answer to your question is 8.28 g of glucose
Explanation:
Data
Glucose (C₆H₁₂O₆) = ?
Ethanol (CH₃CH₂OH)
Carbon dioxide (CO₂) = 2.25 l
Pressure = 1 atm
T = 295°K
Reaction
C₆H₁₂O₆ ⇒ 2C₂H₅OH(l) +2CO₂(g)
- Calculate the number of moles
PV = nRT
Solve for n
Substitution
Simplification
n = 0.092
- Calculate the mass of glucose
1 mol of glucose --------------- 2 moles of carbon dioxide
x --------------- 0.092 moles
x = (0.092 x 1) / 2
x = 0.046 moles of glucose
Molecular weight of glucose = 180 g
180 g of glucose --------------- 1 mol
x g ---------------0.046 moles
x = (0.046 x 180) / 1
x = 8.28 g of glucose
