Answer:
i think the answer is B cus i think of that
Answer:
0.296 J/g°C
Explanation:
Step 1:
Data obtained from the question.
Mass (M) =35g
Heat Absorbed (Q) = 1606 J
Initial temperature (T1) = 10°C
Final temperature (T2) = 165°C
Change in temperature (ΔT) = T2 – T1 = 165°C – 10°C = 155°C
Specific heat capacity (C) =..?
Step 2:
Determination of the specific heat capacity of iron.
Q = MCΔT
C = Q/MΔT
C = 1606 / (35 x 155)
C = 0.296 J/g°C
Therefore, the specific heat capacity of iron is 0.296 J/g°C
Answer: (1 Kilogram = 2.20462 pounds) . There are 2.2046226218 lb in 1 kilogram. To convert kilograms to pounds, multiply your figure by 2.205 for an approximate result. 1 kilogram is also equal to 2 lb and 3.27396195 oz. Working out a rough estimate in your head for converting to pounds and ounces may be tricky - remember that there are 16 ounces in a pound.
Lets name the unknown metal as M. Cation would be M³⁺.
the molecular formula of the compound is M₂(SO₄)₃
the mass of one mole - (molar mass of M x2 + 3 x molar mass of SO₄²⁻)
= 2M + 96 x 3
= 2M + 288
In 1 mol if there's 72.07% of sulphate ,
then 72.07 % corresponds to 288 g
1 % is then - 288/72.07
100 % of the compound - 288/72.07 x 100
molar mass of the compound - 399.6 g/mol
mass of 2M - 399.6 - 288 = 111.6 g
molar mass of M - 111.6 /2 = 55.8 g/mol
the element with molar mass of 55.8 is Fe.
Unknown metal is iron(III) , Fe³⁺
207.217 amu
Work:
203.973 amu *(0.014) = 2.855 amu
205.974 amu *(0.241) = 49.639 amu
206.976 amu *(0.221) = 45.741 amu
207.977 amu *(0.524) = 108.979 amu
2.855 + 49.639 + 45.741 + 108.979 = 207.217 amu
