Answer:
MgBr(aq) + (NH4)3PO4(aq) -------> NH4Br(aq) + Mg3(PO4)2(s)
Explanation:
Answer:
Acid: H^+ Base: OH^- Have a great day.....
Explanation:
Answer:
After 2.0 minutes the concentration of N2O is 0.3325 M
Explanation:
Step 1: Data given
rate = k[N2O]
initial concentration of N2O of 0.50 M
k = 3.4 * 10^-3/s
Step 2: The balanced equation
2N2O(g) → 2 N2(g) + O2(g)
Step 3: Calculate the concentration of N2O after 2.0 minutes
We use the rate law to derive a time dependent equation.
-d[N2O]/dt = k[N2O]
ln[N2O] = -kt + ln[N2O]i
⇒ with k = 3.4 *10^-3 /s
⇒ with t = 2.0 minutes = 120s
⇒ with [N2O]i = initial conc of N2O = 0.50 M
ln[N2O] = -(3.4*10^-3/s)*(120s) + ln(0.5)
ln[N2O] = -1.101
e^(ln[N2O]) = e^(-1.1011)
[N2O} = 0.3325 M
After 2.0 minutes the concentration of N2O is 0.3325 M
Answer:
Density of the copper = 8.94g/cm^3
Student A results = 7.3gm/cm^3 ,9.4 gm/cm^3 , 8.3gm/cm^3
Student B results = 8.4 gm/cm^3 , 8.8 gm/cm^3 , 8gm/cm^3
From the observations we conclude that
Student A's result is accurate but not precise as the trials noted are not close to each other.
Student B's result is accurate and precise as the trials noted are close to each other.
Mean density of student A = 7.3 + 9.4 + 8.3 /3 = 8.33gm/cm^3
Mean density of student B = 8.4 + 8.8 + 8 /3 = 8.4 gm/cm^3
both the densities of A and B are 0.5 away from the actual density.
Answer:
The pressure increases by a factor of four.
Explanation:
Let's consider a gas at a given temperature and pressure (T₁, P₁). The absolute temperature of a gas is increased four times (T₂ = 4 T₁) while maintaining a constant volume. We can assess the effect on the pressure (P₂) by using Gay Lussac's law.
P₁/T₁ = P₂/T₂
P₂ = P₁ × T₂/T₁
P₂ = P₁ × 4 T₁/T₁
P₂ = 4 P₁
The pressure increases by a factor of four.
