Answer:
The correct answer is: pH= 4.70
Explanation:
We use the <em>Henderson-Hasselbach equation</em> in order to calculate the pH of a buffer solution:
![pH= pKa + log \frac{ [conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pH%3D%20pKa%20%2B%20log%20%20%20%5Cfrac%7B%20%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D)
Given:
pKa= 4.90
[conjugate base]= 4.75 mol
[acid]= 7.50 mol
We calculate pH as follows:
pH = 4.90 + log (4.75 mol/7.50 mol) = 4.90 + (-0.20) = 4.70
K₃PO₄<span>(aq) reacts with Al(NO</span>₃)₃(aq) and form AlPO₄(s) and KNO₃(aq) as the products.
Before writing net ionic reaction, following steps should be followed.
Step 1 : Write the balanced chemical equation for the reaction and indicate the state of compound as aq,s,g or l.
K₃PO₄(aq) + Al(NO₃)₃(aq) → AlPO₄(s) + 3KNO₃(aq)
Step 2 : Identify the ionic species which dissolve in water to form ions.
K₃PO₄(aq) → 3K⁺(aq) + PO₄⁻(aq)
Al(NO₃)₃(aq) → Al³⁺(aq) + 3NO₃⁻(aq)
KNO₃(aq) → K⁺(aq) + NO₃⁻(aq)
Step 3 : Write the equation again by using ions.
3K⁺(aq) + PO₄⁻(aq) + Al³⁺(aq) + 3NO₃⁻(aq) → AlPO₄(s) +3K⁺(aq) + 3NO₃⁻(aq)
Step 4 : Identify the similar species in both side of the equation and cut off them.
3K⁺(aq) and 3NO₃⁻(aq) present in both sides. Hence, those ions can be cut off.
Step 5 : Get the final ionic equation.
The net ionic equation for the given reaction is
Al³⁺(aq) + PO₄⁻(aq) → AlPO₄(s)
Answer:
D. As one variable increases, the other variable also increases
CO2 + H2O = H2CO3
<span>
Your equation is already balanced, and the reaction type is: </span><em><u>synthesis.</u></em>
Answer:
(a) The slope of distance-time graph gives the speed of the objects.
As the slope of object (B) is the greatest among all, thus B has the largest speed and hence it is the fastest among all the objects.
(b) : As the curve of the objects do not intersect with each other even once, hence they cannot be at the same point on the road at the same time.
(c) : On the y axis (i.e. distance line), 7 small represents 4 km
∴ 1 small line =
7
4
=0.57 km
Position of C at the instant B passes A is 8 km
(d) ; Position of B at the instant it passes C is 2nd small line above 4 km i.e. 4+2×0.57=5.14 km
∴ Distance traveled by B =5.14−0=5.14 kmV
Explanation:
