Answer:
The specific heat of the alloy 
Explanation:
Mass of an alloy
= 25 gm
Initial temperature
= 100°c = 373 K
Mass of water
= 90 gm
Initial temperature of water
= 25.32 °c = 298.32 K
Final temperature
= 27.18 °c = 300.18 K
From energy balance equation
Heat lost by alloy = Heat gain by water
[
-
] =
(
-
)
25 ×
× ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

This is the specific heat of the alloy.
<span>let x=gallons of current mixture to be drained
and replaced with pure antifreeze.
4-x=gallons of current mixture remaining in the car.</span>
<span>
0.15(4-x)+1.00x=0.50 x 4
0.6-.15x+x=2
0.85x=1.4
x=1.4/0.85 =1.65 gal
Thus, 1.65 gallons of current mixture to be drained and replaced with pure
antifreeze.</span>
Answer:
alright bud lets see hmm.... the answer is a. 90.5kpa
Explanation:
92.3 kPa - 1.82 kPa = 90.5 kPa
keep up your hope also corrected me if a am wrong in anyway! :)
Explanation:
By losing or gaining electrons from its outermost orbit
Answer:
0.225 mol = 0.23 mol to 2 significant figures
Explanation:
Calculate the moles of oxygen needed to produce 0.090 mol of water
The equation of the reaction is given as;
2 C2H2 + 5 O2 --> 4 CO2 + 2 H2O
From the equation of the reaction;
5 mol of O2 produces 2 mol of H2O
x mol of O2 produces 0.090 mol of H2O
5 = 2
x = 0.090
x = 0.090 * 5 / 2
x = 0.225 mol