Answer:
the answer is three protons
Answer:
1)
(500.mL)(.200M)/.150M = 667 mL
667mL - 500mL = 167mL of water is needed
2)
1.0 L = 1000mL
M1 V1 = M2 V2
(1.6 mol/L) (175 mL) = (x)(1000mL)
x = .28M
As we know that
P.E. = mgh
where,
P.E. = Potential energy of the object =?
m= mass of object= 3kg
g= acceleration due to gravity = 9.8 ms^-2
h = height between object and animal = 0 m
Then
P.E. = 3× 9.8 × 0 = 0 Joules or 0J
<em>Have a luvely day!</em>
Answer:
The correct answer is 574.59 grams.
Explanation:
Based on the given information, the number of moles of NH₃ will be,
= 2.50 L × 0.800 mol/L
= 2 mol
The given pH of a buffer is 8.53
pH + pOH = 14.00
pOH = 14.00 - pH
pOH = 14.00 - 8.53
pOH = 5.47
The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,
= -log (1.8 ×10⁻⁵)
= 5.00 - log 1.8
= 5.00 - 0.26
= 4.74
Based on Henderson equation:
pOH = pKb + log ([salt]/[base])
pOH = pKb + [NH₄⁺]/[NH₃]
5.47 = 4.74 + log ([NH₄⁺]/[NH₃])
log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73
[NH₄⁺]/[NH₃] = 10^0.73= 5.37
[NH₄⁺ = 5.37 × 2 mol = 10.74 mol
Now the mass of dry ammonium chloride required is,
mass of NH₄Cl = 10.74 mol × 53.5 g/mol
= 574.59 grams.
