The amount, in mL, of the concentrated acid required, would be 1.1875 mL
<h3>Dilution</h3>
From the dilution equation:
m1v1=m2v2 where m1 and m2 = molarity before and after dilution, and v1 and v2 = volume before and after dilution.
m2 = 0.285M, m1 = 12.0M v2 = 50.0 mL
v1 = m2v2/m1 = 0.285x50/12 = 1.1875 mL
Thus, 1.1875 mL of the acid would be taken and diluted with water up to the 50 mL mark.
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<span>the solvent, hope this helps</span>
60.7 ml is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa.
Explanation:
Data given:
V1 = 75 ml
T1 = 30 Degrees or 273.15 + 30 = 303.15 K
P1 = 91 KPa
V2 =?
P2 = 1 atm or 101.3 KPa
T2 = 273.15 K
At STP the pressure is 1 atm and the temperature is 273.15 K
applying Gas Law:
= 
putting the values in the equation of Gas Law:
V2 = 
V2 = 
V2 = 60.7 ml
at STP the volume of carbon dioxide gas is 60.7 ml.
Answer:
The answer to your question is V = 0.108 L or 108 ml
Explanation:
Data
Volume = ?
mass = 0.405 g
Temperature = 273°K
Pressure = 1 atm
Process
1.- Convert mass of Kr to moles
83.8 g of Kr -------------------- 1 mol
0.405 g ------------------- x
x = (0.405 x 1) / 83.8
x = 0.0048 moles
2.- Use the Ideal gas law to solve this problem
PV = nRT
- Solve for V
V = nRT / P
- Substitution
V = (0.0048)(0.082)(273) / 1
- Simplification
V = 0.108 / 1
- Result
V = 0.108 L
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