Answer:
(1) order = 2
(2) R = K [A]²
Explanation:
Given the reaction:
A--------->Product
The rate constant relation for the reaction is given as:
R(i) = K [A]............(*)
Where R(I) is rate constant at different concentration of A.
Taking the rate constant as R1, R2 and R3 for the different concentrations respectively. Then the following equations results
0.011 = K [0.15] ⁿ.........(1)
0.044 = K [0.30]ⁿ .......(2)
0.177 = K [0.60]ⁿ .........(3)
Dividing (2) by (1) and (3) by (1)
Gives:
0.044/0.011 = [0.3/0.15]ⁿ
4 = 2ⁿ; 2² = 2ⁿ; n = 2
Similarly
0.177/0.011 = [0.60/0.15]ⁿ
16.09 = 4ⁿ
16.09 = 16 (approximately)
4² = 4ⁿ ; n = 2
Hence the order of the reaction is 2.
The rate law is R = K [A]²
Answer:
4.62
so 5
the ratio is 2 na chlorates for 3 O2 so multiply 7 by 2/3
Explanation:
Formula:
f = C/λ
Where,
λ (Lambda) = Wavelength in meters
c = Speed of Light (299,792,458 m/s)
f = Frequency (MHz)
answer= 693964023.148148 MHz
Energy absorbed by Iron block E (iron) = 460.5 J
Energy absorbed by Copper block E (Copper) = 376.8 J
<u>Explanation:</u>
To find the heat absorbed, we can use the formula as,
q = m c ΔT
Here, Mass = m = 10 g = 0.01 kg
ΔT = change in temperature = 400 - 300 = 100 K = 100 - 273 = -173 °C
c = specific heat capacity
c for iron = 460.5 J/kg K
c for copper = 376.8 J/kg K
Plugin the values in the above equation, we will get,
q (iron) = 0.01 kg × 460.5 J/kg K × 100 K
= 460.5 J
q (copper) = 0.01 kg × 376.8 J/kg K × 100 K
= 376.8 J
Answer:
(A) 4.616 * 10⁻⁶ M
(B) 0.576 mg CuSO₄·5H₂O
Explanation:
- The molar weight of CuSO₄·5H₂O is:
63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol
- The molarity of the first solution is:
(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M
The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.
- Now we use the dilution factor in order to calculate the molarity in the second solution:
(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M
To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:
- 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
- 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O
