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Answer:
Heat transfer during melting of ice plays greater role in cooling of liquid water.
Explanation:
Temperature of ice = -10 °c
Temperature of water = 0 °c
When ice cube is dipped in to the water.the heat transfer
Q = m c ΔT
⇒ Q = 1 × 2.01 × 10
⇒ Q = 20.1 KJ
Heat transfer during melting of ice = latent heat of ice
Latent heat of ice = 334 KJ
⇒ = 334 KJ
Heat transfer during melting of ice is greater value than heat transfer during warming of ice from -10°C to 0°C.
Thus heat transfer during melting of ice plays greater role in cooling of liquid water.
The question requires us to use the dilution formula , where
and
are the stock concentration and volume respectively, then
and
are the dilute concentration and volume respectively.
a. = 1.5 M KCI,
=0.20M KCl,
=250ml KCl
b.
To prepare the solution 33.3ml of 1.5M KCl is diluted to a total final volume of 250ml.
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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