4.00 moles of sodium has a mass of 91.96 grams.
To find an element's molar mass, you can use its atomic weight.
Answer:
The answer to your question is: Initial temperature of copper = 67.1°C
Explanation:
Data
mass Copper = 248 g
volume Water = 390 ml
T1 water = 22.6°C
T2 = 39.9°C
T1 copper = ?
Specific heat water = 1 cal/g°C
Specific heat copper = 0.092 cal/g°C
Formula copper water
Heat is negative for copper because it releases heat
- mCp(T2 - T1) = mCp(T2 - T1)
- (248)(39.9 - T1) = 390 (1)((39.9 - 22.6) Substitution
-9895.2 + 248T1 = 390(17.3) Simplification
-9895.2 + 248T1 = 6747
248 T1 = 6747 + 9895.2
248 T1 = 16642.2
T1 = 16642.2 / 248
T1 = 67.1 °C Result
Answer:
The answer is Frost Point.
Explanation:
The temperature to which the air must be cooled, with constant pressure, to reach saturation (in relation to liquid water), is called the dew point. The dew point gives a measure of the water vapor content in the air. The higher, the greater the concentration of water vapor in the air. However, when cooling produces saturation at a temperature of 0 ° C or less, the temperature is called a frost point. The water vapor is deposited as frost on a surface whose temperature is below the dew point.
Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %