Answer:
Amplitude is measured from the center line to the highest point in the waves.
Wavelength is the distance between one wave to the other from the highest point.
Frequency is the rate of the waves.
Answer:
0.435 M
Explanation:
In case of dilution , the following formula can be used -
M₁V₁ = M₂V₂
where ,
M₁ = initial concentration ,
V₁ = initial volume ,
M₂ = final concentration , i.e. , concentration after dilution ,
V₂ = final volume .
from , the question ,
M₁ = 0.725 M
V₁ = 300 mL
M₂ = ?
V₂ = 300 mL + 200 mL = 500 mL
Since, the final volume of solution would be the summation of the initial and final volume.
Using the above formula , the molarity of the final solution after dilution , can be calculated as ,
M₁V₁ = M₂V₂
0.725 M * 300 mL = M₂ * 500mL
M₂ = 0.435 M
Answer:
Molar concentration of HCl = 0.279 M
Explanation:
Given data:
Volume of NaOH = 31.00 mL ( 31/1000 = 0.031 L)
Molarity of NaOH = 0.0900 M
Volume of HCl = 10.00 mL (0.01 L)
Molar concentration of HCl = ?
Solution:
Chemical equation:
NaOH + HCl → NaCl + H₂O
Number of moles of NaOH:
Number of moles = Molarity × volume in L
Number of moles = 0.0900 mol/L × 0.031 L
Number of moles = 0.00279 mol
For one mole of HCl one mole of NaOH is required.
Number of moles of HCl = 0.00279 mol
Molar concentration of HCl = 0.00279 mol / 0.01 L
Molar concentration of HCl = 0.279 M
<u>Given:</u>
Mass of ice = mass of water = 5.50 kg = 5500 g
Temperature of ice = -20 C
Temperature of water = 75 C
<u>To determine:</u>
Mass of propane required
<u>Explanation:</u>
Heat required to change from ice to water under the specified conditions is:-
q = q(-20 C to 0 C) + q(fusion) + q (0 C to 75 C)
= m*c(ice)*ΔT(ice) + m*ΔHfusion + m*c(water)*ΔT(water)
= 5500[2.10(0-(-20)) + 334 + 4.18(75-0)] = 3792 kJ
The enthalpy change for the combustion of propane is -2220 kJ/mol
Therefore, the number of moles of propane corresponding to the required energy of 3792 kJ = 1 mole * 3792 kJ/2220 kJ = 1.708 moles of propane
Molar mass of propane = 44 g/mol
Mass of propane required = 1.708 moles * 44 g/mol = 75.15 g
Ans: 75.15 grams of propane must be combusted.
Answer:
absolutely B because Al2O3 could react with axit and bazo
Explanation: