Answer:
B. How much energy it takes to heat a substance
Explanation:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
The substances with higher value of specific heat capacity require more heat to raise the temperature by one degree as compared the substances having low value of specific heat capacity. For example,
The specific heat capacity of oil is 1.57 j/g. K and for water is 4.18 j/g.K. So, water take a time to increase its temperature by one degree by absorbing more heat while oil will heat up faster by absorbing less amount of heat.
Consider that both oil and water have same mass of 5g and change in temperature is 15 K. Thus amount of heat thy absorbed to raise the temperature is,
For oil:
Q = m.c. ΔT
Q = 5 g× 1.67 j/g K × 15 K
Q = 125.25 j
For water:
Q = m.c. ΔT
Q = 5 g× 4.18 j/g K × 15 K
Q = 313.5 j
we can observe that water require more heat which is 313.5 j to increase its temperature.
The answer is:
C) Sound is a mechanical wave and needs a medium to travel through while light is an electromagnetic wave.
Answer:
166 g/mol
Explanation:
Step 1: Write the neutralization reaction
H₂A + 2 NaOH ⇒ Na₂A + 2 H₂O
Step 2: Calculate the reacting moles of NaOH
48.3 mL of 0.0700 M NaOH react.
0.0483 L × 0.0700 mol/L = 3.38 × 10⁻³ mol
Step 3: Calculate the reacting moles of H₂A
The molar ratio of H₂A to NaOH is 1:2. The reacting moles of H₂A are 1/2 × 3.38 × 10⁻³ mol = 1.69 × 10⁻³ mol.
Step 4: Calculate the molar mass of H₂A
1.69 × 10⁻³ moles of H₂A have a mass of 0.281 g. The molar mass of H₂A is:
M = 0.281 g / 1.69 × 10⁻³ mol = 166 g/mol
Answer:
1.08mol
Explanation:
moles = reacting mass/ molecular weight
reacting mass = 80.0g molecular weight of Ca[OH]2= 40 + 2(16 +1) = 74g/mol
mole = 80.0/74 = 1.08mol
The Ideal Gas Law states that pressure (P) × volume (V) is equal to the # of moles (n) of the gas × a constant (R) × temperature (T), such that the equation is:
PV = nRT
At standard temp and pressure (STP), the T is 0°C or 273.15K, the P is 1 atm or 760 torr, and the R constant is 0.0821. Therefore the equation, solved for V becomes: V = nRT/P, or V = n(0.0821)(273)/1, so that it reduces to V = 22.4 Liters, when n = 1 mole.
So the V of any gas at STP is 22.4 L / mole
