Protons are positively charged neutrons are neutrally charged amd electrons are negatively charged. Therefore only the protons in this instance are charged
Answer: 0.151
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Given:
Putting in the values we get:
Thus the rate of appearance of is 0.151
Answer:
0.1
Explanation:
We must first put down the equation of the reaction in order to guide our solution of the question.
2HNO3(aq) + Sr(OH)2(aq) -------> Sr(NO3)2(aq) + 2H2O(l)
Now from the question, the following were given;
Concentration of acid CA= ??????
Concentration of base CB= 0.299M
Volume of acid VA= 17.8ml
Volume of base VB= 24.7ml
Number of moles of acid NA= 2
Number of moles of base NB= 1
From;
CAVA/CBVB= NA/NB
CAVANB= CBVBNA
CA= CBVBNA/VANB
SUBSTITUTING VALUES;
CA= 0.299 × 24.7 ×2 / 17.8×1
CA= 0.8298 M
But;
pH= -log[H^+]
[H^+] = 0.8298 M
pH= -log[0.8298 M]
pH= 0.1
Answer:
0.03g/mL
Explanation:
Given parameters include:
Five μL of a 10-to-1 dilution of a sample; This implies the Volume of dilute sample is given as 5 μL
Dilution factor = 10-to-1
The absorbance at 595 nm was 0.78
Mass of the diluted sample = 0.015 mg
We need to first determine the concentration of the diluted sample which is required in calculating the protein concentration of the original solution.
So, to determine the concentration of the diluted sample, we have:
concentration of diluted sample =
= (where ∝ was use in place of μ in the expressed fraction)
= 0.003 mg/μL
The dilution of the sample is from 10-to-1 indicating that the original concentration is ten times higher; as such the protein concentration of the original solution can be calculated as:
protein concentration of the original solution = 10 × concentration of the diluted sample.
= 10 × 0.003 mg/μL
= 0.03 mg/μL
= 0.03g/mL
Hence, the protein concentration of the original solution is known to be 0.03g/mL
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hope this helps!