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hoa [83]
3 years ago
8

The "correct" mass of the cylinder had been previously determined to be 27.20 grams. What was the percent error on his first wei

ghing if Bart found the mass to be 25.01 g. Show your work.

Chemistry
1 answer:
Paraphin [41]3 years ago
5 0

Answer:

Percent error =  - 8.1 %

Explanation:

Given data:

Actual mass = 27.20 g

Theoretical mass = 25.01 g

Percent error = ?

Solution:

Formula:

Percent error = theoretical value- actual value/ actual value × 100

Now we will put the values in formula.

Percent error =   25.01 - 27.2 / 27.2 × 100

Percent error = -2.19 / 27.2 × 100

Percent error = -0.081  × 100

Percent error =  - 8.1 %

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If you have access to stock solutions of 1.00 M H3PO4, 1.00 M of HCl, and 1.00 M NaOH solution, (and distilled water of course),
garri49 [273]

Answer:

0.10L of 1.00M of H₃PO₄ and 0.1613L of 1.00M NaOH

Explanation:

The pKa's of phosphoric acid are:

H₃PO₄/H₂PO₄⁻ = 2.1

H₂PO₄⁻/HPO₄²⁻ = 7.2

HPO₄²⁻/PO₄³⁻ = 12.0

To make a buffer with pH 9.40 we need to convert all H₃PO₄ to H₂PO₄⁻ and an amount of H₂PO₄⁻ to HPO₄²⁻

To have a 50mM solution of phosphoures we need:

2L * (0.050mol / L) = 0.10 moles of H₃PO₄

0.10 mol * (1L / mol) = 0.10L of 1.00M of H3PO4

To convert the H₃PO₄ to H₂PO₄⁻ and to HPO₄²⁻ must be added NaOH, thus:

H₃PO₄ + NaOH → H₂PO₄⁻ + H₂O + Na⁺

H₂PO₄⁻ + NaOH → HPO₄²⁻ + H₂O + Na⁺

Using H-H equation we can find the amount of NaOH added:

pH = pKa + log [A⁻] / [HA] <em>(1)</em>

<em>Where [A-] is conjugate base, HPO₄²⁻ and [HA] is weak acid, H₂PO₄⁻</em>

<em>pH = 7.40</em>

<em>pKa = 7.20</em>

[A-] + [HA] = 0.10moles <em>(2)</em>

Replacing (2) in (1):

7.40 = 7.20 + log 0.10mol - [HA] / [HA]

0.2 = log 0.10mol - [HA] / [HA]

1.5849 = 0.10mol - [HA] / [HA]

1.5849 [HA] = 0.10mol - [HA]

2.5849[HA] = 0.10mol

[HA] = 0.0387 moles = H₂PO₄⁻ moles

That means moles of HPO₄²⁻ are 0.10mol - 0.0387moles = 0.0613 moles

The moles of NaOH needed to convert all H₃PO₄ in H₂PO₄⁻ are 0.10 moles

And moles needed to obtain 0.0613 moles of HPO₄²⁻ are 0.0613 moles

Total moles of NaOH are 0.1613moles * (1L / 1mol) = 0.1613L of 1.00M NaOH

Then, you need to dilute both solutions to 2.00L with distilled water.

4 0
2 years ago
At 73.0 ∘c , what is the maximum value of the reaction quotient, q, needed to produce a non-negative e value for the reaction so
damaskus [11]
Here we will use the general formula of Nernst equation:

Ecell = E°Cell - [(RT/nF)] *㏑Q

when E cell is cell potential at non - standard state conditions

E°Cell is standard state cell potential = - 0.87 V

and R is a constant = 8.314 J/mol K

and T is the temperature in Kelvin = 73 + 273 = 346 K

and F is Faraday's constant = 96485 C/mole

and n is the number of moles of electron transferred in the reaction=2  

and Q is the reaction quotient for the reaction 
SO42-2(aq) + 4H+(aq) +2Br-(aq) ↔  Br2(aq) + SO2(g) +2H2O(l)

so by substitution :

0 = -0.87 - [(8.314*346K)/(2* 96485)*㏑Q      → solve for Q 


∴ Q = 4.5 x 10^-26 
6 0
2 years ago
Can someone help me please .
lyudmila [28]

Sure. Whith what do you need help?

4 0
3 years ago
Combustion of 9.511 grams of c4h10 will yield ____ grams of CO2
Flauer [41]

Answer:

\boxed{28.81}

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:      58.12                   44.01

           2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

m/g:     9.511

1. Moles of C₄H₁₀

\text{Moles of C$_{4}$H$_{10} $} = \text{ 9.511 g C$_{4}$H$_{10} $} \times \dfrac{\text{1 mol C$_{4}$H$_{10} $}}{\text{ 58.12 g C$_{4}$H$_{10} $}} = \text{0.1636 mol C$_{4}$H$_{10}$}

2. Moles of CO₂

The molar ratio is 8 mol CO₂:2 mol C₄H₁₀

\text{Moles of CO}_{2} =\text{0.1636 mol C$_{4}$H$_{10} $} \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \text{0.6546 mol CO}_{2}

3. Mass of CO₂

\text{Mass of CO}_{2} = \text{0.6546 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{28.81 g CO}_{2}\\\\\text{The combustion will form $\boxed{\textbf{28.81 g CO}_{2}}$}

8 0
2 years ago
Which is an example of potential energy?
Setler79 [48]
Probably C hope this helps
5 0
2 years ago
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