<h3>
Answer:</h3>
0.424 J/g °C
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Chemistry</u>
<u>Thermochemistry</u>
Specific Heat Formula: q = mcΔT
- q is heat (in Joules)
- m is mass (in grams)
- c is specific heat (in J/g °C)
- ΔT is change in temperature
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] m = 38.8 g
[Given] q = 181 J
[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C
[Solve] c
<u>Step 2: Solve for Specific Heat</u>
- Substitute in variables [Specific Heat Formula]: 181 J = (38.8 g)c(11.0 °C)
- Multiply: 181 J = (426.8 g °C)c
- [Division Property of Equality] Isolate <em>c</em>: 0.424086 J/g °C = c
- Rewrite: c = 0.424086 J/g °C
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.424086 J/g °C ≈ 0.424 J/g °C
Gravity is the force that draws water back to the earth in the forms of rain, sleet and, snow.
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+ + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:
PH= -㏒[H+]
= -㏒(4.81x10^-7) = 6.32
