Answer:
The final temperature is 348.024°C.
Explanation:
Given data:
Specific heat of copper = 0.385 j/g.°C
Energy absorbed = 7.67 Kj (7.67×1000 = 7670 j)
Mass of copper = 62.0 g
Initial temperature T1 = 26.7°C
Final temperature T2 = ?
Solution:
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
ΔT = T2 - T1
Q = m.c. ΔT
7670 J = 62.0 g × 0.385 j/g °C ×( T2- 26.7 °C
)
7670 J = 23.87 j.°C ×( T2- 26.7 °C
)
7670 J / 23.87 j/°C = T2- 26.7 °C
T2- 26.7 °C = 321.324°C
T2 = 321.324°C + 26.7 °C
T2 = 348.024°C
The final temperature is 348.024°C.
Answer:
The answer is 465.6 mg of MgI₂ to be added.
Explanation:
We find the mole of ion I⁻ in the final solution
C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol
But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.
So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.
Hence, the weight of MgI₂ must be added is
Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg
Answer:
B. Measures of central tendency
Explanation:
Mean, median and mode are best described as measures of central tendency of a given data set.
Mean is the average of the samples given
Mode is the data point with the most frequent occurrence
Median is the data point that lies in the middle
- All these parameters tells us how far a data point is from the middle or how close they are.
Answer:
The answer is 0.0698 M
Explanation:
The concentration was prepared by a serial dilution method.
The formula for the preparation I M1V1 = M2V2
M1= the concentration of the stock solution = 0.171 M
V1= volume of the stock solution taken = 200 mL
M2 = the concentration produced
V2 = the volume of the solution produced = 940 mL
Substitute these values in the formula
0.171 × 200 = 490 × M2
34.2 = 490 × M2
Make M2 the subject of the formula
M2 = 34.2/490
M2 = 0.069795
M2 = 0.0698 M ( 3 s.f)
The concentration of the Chemist's working solution to 3 significant figures is 0.0698M