<span>Heat
gained or absorbed in a system can be calculated by multiplying the given mass to the
specific heat capacity of the substance and the temperature difference. The heat capacity of aluminum at 25 degrees celsius is 0.9 J/g-C. It is
expressed as follows:</span><span>
Heat = mC(T2-T1)
5800 J = 152000(0.90)(</span>ΔT)
ΔT = 0.42 °C change in temperature
Answer:
Correct, because B it is reported to the nearest miligram
Explanation:
4.6 rounded up is 5
Answer:- 
Solution:- It is a volume unit conversion problem where we are asked to convert the volume from 
to microliters.
We know that:
= 1 mL
and, 
Let's use these conversions factors for the desired conversion using dimensional as:
=
So, the answer is .
Answer:
5
Explanation:
Given that the formula is;
1/λ= R(1/nf^2 - 1/ni^2)
λ = 93.7 nm or 93.7 * 10^-9 m
R= 1.097 * 10^7 m-1
nf = ?
ni = 1
From;
ΔE = hc/λ
ΔE = 6.63 * 10^-34 * 3* 10^8/93.7 * 10^-9
ΔE = 21 * 10^-19 J
ΔE = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)
21 * 10^-19 J = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)
21 * 10^-19/-2.18 * 10^-18 = (1/nf^2 - 1/1^2)
-0.963 = (1/nf^2 - 1)
-0.963 + 1 = 1/nf^2
0.037 = 1/nf^2
nf^2 = (0.037)^-1
nf^2 = 27
nf = 5
1.04gK*1molK/39.01g K= 0.0267 mol K
0.70gCr*1mol/52.0g Cr = <span>0.0135 mol Cr
0.86 gO* 1 mol/16.0 g O = 0.0538 mol O
</span>0.0267 mol K/0.0135 = 2 mol K
0.0135 mol Cr /0.0135= 1 mol Cr
0.0538 mol O/0.035= 4 mol Cr
K2CrO4
