Explanation:
Cadmium, nickel, chromium, and silver are sometimes used as protective platings. Metals have a wide range of corrosion resistance. The most active metals (i.e., those that tend to lose electrons easily)--such as magnesium and aluminum--corrode easily and are listed at the top of Table 2-1.
2 ICl + H2 ----> I2 + 2 HCl
as given that rate is first order with respect to ICl and second order with respect to H2
The rate law will be
Rate = K [ICl] [ H2]^2
b) Given that K = 2.01 M^-2 s^-1
Concentrations are
[ICl] = 0.273 m and [H2] = 0.217 m
Therefore rate = 2.01 X (0.273)(0.217)^2 = 0.0258 M / s
Answer:
V₂ ≈416.7 mL
Explanation:
This question asks us to find the volume, given another volume and 2 temperatures in Kelvin. Based on this information, we must be using Charles's Law and the formula. Remember, his law states the volume of a gas is proportional to the temperature.
where V₁ and V₂ are the first and second volumes, and T₁ and T₂ are the first and second temperature.
The balloon has a volume of 600 milliliters and a temperature of 360 K, but the temperature then drops to 250 K. So,
- V₁= 600 mL
- T₁= 360 K
- T₂= 250 K
Substitute the values into the formula.
- 600 mL /360 K = V₂ / 250 K
Since we are solving for the second volume when the temperature is 250 K, we have to isolate the variable V₂. It is being divided by 250 K. The inverse o division is multiplication, so we multiply both sides by 250 K.
- 250 K * 600 mL /360 K = V₂ / 250 K * 250 K
- 250 K * 600 mL/360 K = V₂
The units of Kelvin cancel, so we are left with the units of mL.
- 250 * 600 mL/360=V₂
- 416.666666667 mL= V₂
Let's round to the nearest tenth. The 6 in the hundredth place tells us to round to 6 to a 7.
The volume of the balloon at 250 K is approximately 416.7 milliliters.
a. 0.137
b. 0.0274
c. 1.5892 g
d. 0.1781
e. 5.6992 g
<h3>Further explanation</h3>
Given
Reaction
2 C4H10 + 13O2 -------> 8CO2 + 10H2O
2.46 g of water
Required
moles and mass
Solution
a. moles of water :
2.46 g : 18 g/mol = 0.137
b. moles of butane :
= 2/10 x mol water
= 2/10 x 0.137
= 0.0274
c. mass of butane :
= 0.0274 x 58 g/mol
= 1.5892 g
d. moles of oxygen :
= 13/2 x mol butane
= 13/2 x 0.0274
= 0.1781
e. mass of oxygen :
= 0.1781 x 32 g/mol
= 5.6992 g
