Equation is as follow,
<span> 2 AgNO</span>₃<span> + MgBr</span>₂<span> </span>→ <span>2 AgBr + Mg(NO</span>₃<span>)</span>₂
According to eq.
339.74 g (2 moles) AgNO₃ produces = 375.54 g (2 moles) of AgBr
So,
22.5 g AgNO₃ will produce = X g of AgBr
Solving for X,
X = (22.5 g × 375.54 g) ÷ 339.74 g
X = 24.87 g of AgBr
First step is to balance the reaction equation. Hence we get
P4 + 5 O2 => 2 P2O5
Second, we calculate the amounts we start with
P4: 112 g = 112 g/ 124 g/mol – 0.903 mol
O2: 112 g = 112 g / 32 g/mol = 3.5 mol
Lastly, we calculate the amount of P2O5 produced.
2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4
mol of P2O5.
This is 1.4 * (31*2 + 16*5) = 198.8 g
Answer:
Kinetic Energy and Potential Energy
Explanation:
When an object is in motion(ie.moving), it has kinetic energy. Energy stored in an object due to position and state, is called potential energy.
The first order rate law has the form: -d[A]/dt = k[A] where, A refers to cyclopropane. We integrate this expression in order to arrive at an equation that expresses concentration as a function of time. After integration, the first order rate equation becomes:
ln [A] = -kt + ln [A]_o, where,
k is the rate constant
t is the time of the reaction
[A] is the concentration of the species at the given time
[A]_o is the initial concentration of the species
For this problem, we simply substitute the known values to the equation as in:
ln[A] = -(6.7 x 10⁻⁴ s⁻¹)(644 s) + ln (1.33 M)
We then determine that the final concentration of cyclopropane after 644 s is 0.86 M.
The unbalanced Equation is
electricity
KI (aq) + H2O (l) -------------> H2 (g) + I2(g) + KOH(aq)
