Answer:
The specific heat capacity of silver is 0.24 j/g.°C.
Explanation:
Given data:
Mass of sample = 55.00 g
increase of temperature ΔT= 15.0 °C
Heat absorbed = 193.9 J
Specific heat capacity of silver = ?
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
Now we will put the values in formula.
193.9 J = 55.00 g × c ×15.0 °C
193.9 J = 825 g.°C × c
c = 193.9 J / 825 g.°C
c= 0.24 /g.°C
The specific heat capacity of silver is 0.24 j/g.°C.
<em>Answer:</em> 8 (feet)
<em>Explanation:</em>
24 inches = 2 feet
48 inches = 4 feet
12 inches = 1 foot
To find volume you do Base * Width * Height
2*4*1 = 8
Hope this helps!
2.3*10^-2 M/s IS THE ANWSER<span>
</span>
Solution:
Benzoic acid is C6H5COOH
In finding pH
C6H5COOH(aq) <=> C6H5COO^- + H^+ pKa = 4.19, pKa = -logKa so Ka = 10^(-4.19)
Ka = 6.45 x 10^-6
[C6H5COO^-] = x = [H^+]; [C6H5COOH] = 0.5 - x (we are able to make an estimate of [C6H5COOH] = 0.5.
Ka = [H^+][C6H5COO^-]/[C6H5COOH] = x^2/(0.5 - x) = 6.45 x 10^-6
Now,
According to the quadratic equation. x^2 = 3.23 x 10^-5 - 6.45 x 10^-6x
x^2 + (6.45 x 10^-6)x - 3.23 x 10^-5 = 0
enter a = 1, b = 0.00000645, c = 0.0000323
x = 5.68 x 10^-3 = 0.00568 M expression is [C6H5COOH] = 0.5 M is the correct answer.
[H^+] = 0.00568 M, so pH = -log(0.00568 M ) = 2.25
This is the required solution.
The values of N, log N and ln N are all worked out below.
We have the number N as 8.99
hence;
1) lnN = ln(8.99) = 2.196
log N = log (8.99) = 0.954
2) If we have ln N as 3.949
N = e^3.949 = 51.88
So;
log N = log (51.88) = 1.715
3) log N = 0.134
N = Antilog (0.134 ) = 1.36
ln N = ln (1.36) = 0.307
Learn more: brainly.com/question/4210362
