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3 years ago

A volume is measured experimentally as 4.26mL. what is the percentage error, given that the correct value is 4.15mL

2 answers:

8 0

% error = | accepted value - experimental value | / accepted value × 100

Step one- | 4.15mL - 4.26mL | = 0.11

Step two- 0.11 / 4.15 = 0.027

Step three- -0.027 × 100 = 2.7%

kherson [118]3 years ago

5 0

<u>Answer:</u> The percentage error is 2.65 %.

<u>Explanation:</u>

To calculate the percentage error, we use the equation:

$\%\text{ error}=\frac{|\text{Experimental value - Accepted value}|}{\text{Accepted value}}\times 100$

We are given:

Experimental value of volume = 4.26 mL

Accepted value of volume = 4.15 mL

Putting values in above equation, we get:

$\%\text{ error}=\frac{|4.26-4.15|}{4.15}\times 100\\\\\%\text{ error}=2.65\%$

Hence, the percentage error is 2.65 %.

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Calculate the feed ratio of adipic acid and hexamethylene diamine that should be employed to obtain a polyamide of approximately

artcher [175]

Answer:

r= 0.9949 (For 15,000)

r=0.995 (For 19,000)

Explanation:

We know that

Molecular weight of hexamethylene diamine = 116.21 g/mol

Molecular weight of adipic acid = 146.14 g/mol

Molecular weight of water = 18.016 g/mol

As we know that when adipic acid and hexamethylene diamine react then nylon 6, 6 comes out as the final product and release 2 molecule of water.

$M_{repeat}=146.14+166.21-2\times 18.106\ g/mol$

$M_{repeat}=226.32\ g/mol$

Mo= 226.32/2 =113.16 g/mol

$M_n=X_nM_o$

Given that

Mn= 15,000 g/mol

15,000 = Xn x 113.16

Xn = 132.55

Now by using Carothers equation we know that

$X_n=\dfrac{1+r}{1+r-2rp}$

$132.55=\dfrac{1+r}{1+r-2\times 0.99r}$

By calculating we get

r= 0.9949

For 19,000

19,000 = Xn x 113.16

Xn = 167.99

By calculating in same process given above we get

r=0.995

3 0

3 years ago

Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the f

julia-pushkina [17]

Explanation:

As per Brønsted-Lowry concept of acids and bases, chemical species which donate proton are called Brønsted-Lowry acids.

The chemical species which accept proton are called Brønsted-Lowry base.

(a) $HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-$

$HNO_3$ is Bronsted lowry acid and $NO_3^-$ is its conjugate base.

$H_2O$ is Bronsted lowry base and $H_3O^+$ is its conjugate acid.

(b)

$CN^- + H_2O \rightarrow HCN + OH^-$

$CN^-$ is Bronsted lowry base and HCN is its conjugate acid.

$H_2O$ is Bronsted lowry acid and $OH^-$ is its conjugate base.

(c)

$H_2SO_4 + Cl^- \rightarrow HCl + HSO_4^-$

$H_2SO_4$ is Bronsted lowry acid and $HSO_4^-$ is its conjugate base.

Cl^- is Bronsted lowry base and HCl is its conjugate acid.

(d)

$HSO_4^-+OH^- \rightarrow SO_4^{2-}+H_2O$

$HSO_4^-$ is Bronsted lowry acid and $SO_4^{2-}$ is its conjugate base.

OH^- is Bronsted lowry base and $H_2O$ is its conjugate acid.

(e)

$O_{2-}+H_2O \rightarrow 2OH^-$

$O_{2-}$ is Bronsted lowry base and OH- is its conjugate acid.

$H_2O$ is Bronsted lowry acid and OH- is its conjugate base.

6 0

3 years ago

Two easy uses of mixture

lyudmila [28]

Explanation:

it helps to make juices.

It helps to make concentrated acid into dilute acid.

4 0

3 years ago

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Determine the mass of Al(C2H3O2)3 that contains 2.63 Χ 1024 atoms of oxygen.

sergejj [24]

The molecular formula of compound is $Al(C_{2}H_{3}O_{2})_{3}$ .

Since, in 1 mole of $Al(C_{2}H_{3}O_{2})_{3}$ there are $6.023\times 10^{23}$ atoms of $Al(C_{2}H_{3}O_{2})_{3}$ .

Thus, according to molecular formula, in 1 mole of $Al(C_{2}H_{3}O_{2})_{3}$ there are $6\times 6.023\times 10^{23}=3.61\times 10^{24}$ atoms of oxygen atoms.