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kipiarov [429]
3 years ago
6

A student measured the specific heat of water to be 4.29 J/gC. The literature value of the specific heat of water is 4.18J/gC. H

ey What was the students percent error?
Chemistry
1 answer:
natita [175]3 years ago
6 0

Answer:

2.44 %

Explanation:

Given that,

The calculated value of specific heat of water to be 4.29 J/g°C

The actual value of the specific heat of water is 4.18J/g°C

We need to find the percentage error of the student

\text{Percentage error}=\dfrac{|\text{actual value-calculated value}|}{\text{actual value}}\times 100\\\\\%=\dfrac{4.18-4.29}{4.49}\times 100\\\\\%=2.44

Hence, the student's percent error is 2.44 %.

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"You measure 48.9 mL of a solution of sulfuric acid with an unknown concentration, and carefully titrate this solution using a 1
Blizzard [7]

Answer:

C= 0.532M

Explanation:

The equation of reaction is

H2SO4 + 2KOH = K2SO4+ H2O

nA= 1, nB= 2, CA= ?, VA= 48.9ml, CB= 1.5M, VB= 34.7ml

Applying

CAVA/CBVB = nA/nB

(CA× 48.9)/(1.5×34.7)= 1/2

Simplify

CA= 0.532M

6 0
3 years ago
If you have 547.3 grams of Ni2O, how many molecules would be present?
jasenka [17]

Answer: 8.830418848725065

Explanation:

8.830418848725065

5 0
3 years ago
Question 11. Identify the reducing agent <br> Sn+2 + AG 0 —&gt; Sn0 + Ag+
Temka [501]

Answer:

Ag 0 is the reducing agent.

Explanation:

Reducing -> gaining electrons

Oxidizing -> losing electrons

Ag lost electrons (became more positive) since it went from a 0 charge to a +1 charge. Therefore it was oxidized. Ag+ is the oxidized product. Reactants that create an oxidized product are called reducing agents. This would make Ag 0 the reducing agent in this reaction.

4 0
2 years ago
calculate how much acid (acetic acid) and how much conjugate base (sodium acetate) must be used to make 500ml of a 0.8m acetate
kirza4 [7]

For the desired pH of 5.76, 0.365 mol of acetate and 0.035 mol of acid are to be added

let the concentration of acetate be x

then the concentration of acid will be (0.8 - x)

pKa of acetate buffer = 4.76

pH = pKa + log([acetate]/[acid])

⇒4.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 0

⇒x/(0.8-x) = 1

⇒x = 0.4

Therefore

[acetate] = x = 0.4

[acid] = 0.8-x =0.4 M

number of mol = concentration *(volume in mL)

number of mol of acetate = 0.4*0.5

= 0.20 mol

number of mol acid = 0.4*0.5

= 0.20 mol

when desired pH = 5.76

pH = pKa + log([acetate]/[acid])

⇒5.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 1

⇒x/(0.8-x) = 10

⇒x = 8 - 10x

⇒x = 8/11

⇒x= 0.73

[acetate] = x= 0.73

[acid] = 0.8-x = 0.07 M

number of mol = concentration * (volume in mL)

number of mol acetate to be added = 0.73*0.5 = 0.365 mol

number of mol acid to be added = 0.07*0.5 = 0.035 mol

Problem based on acetic acid required to maintain a certain pH

brainly.com/question/9240031

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4 0
1 year ago
Why are valence electrons important?
Usimov [2.4K]
They are the outer layer of the electron layers.
5 0
3 years ago
Read 2 more answers
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