Answer:
0.025 g C6H12O6
Explanation:
ppm = (g solute/ g solution)* 10^6
g solute= (ppm * g solution)/ 10^6
g solute = (250 ppm * 100 g)/10^6
g solute=0.025 g C6H12O6
The balanced equation would be 
<h3>Electrochemical equations</h3>
Zn reacts with Cu solution according to the following equation:

In the reaction,
is reduced according to the following: 
While Zn is oxidized according to the following: 
Thus, giving the overall equation of; 
More oxidation-reduction equations can be found here: brainly.com/question/13699873
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x = 1.01
Explanation:
Given equation:
y = 1.2345x – 0.6789
y = 0.570
Problem:
Solving for x
The variables in this equation are y and x
They can take up any value since they are variables.
Since we have been given y = 0.570
y = 1.2345x – 0.6789
To solve for x, we simply substitute for y in the equation;
since y = 0.570
0.57 = 1.2345x – 0.6789
add 0.6789 to both sides;
0.57 + 0.6789 = 1.2345x – 0.6789 + 0.6789
1.2489 = 1.2345x
Divide both sides by 1.2345
x = 1.01
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I think that it is qualitative data
Answer:
Q = 1360.248 j
Explanation:
Given data:
Mass of brass = 298.3 g
Initial temperature = 30.0°C
Final temperature = 150°C
Specific heat capacity of brass = 0.038 J/g.°C
Heat absorbed = ?
SOLUTION:
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 = 150°C - 30.0°C
ΔT = 120°C
Q = 298.3 g × 0.038 J/g.°C × 120°C
Q = 1360.248 j