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Free_Kalibri [48]
3 years ago
6

Determine the percent error if a student determines that the mass of a block is 45.6 grams and the actual mass is 46.7g.

Chemistry
2 answers:
Charra [1.4K]3 years ago
3 0

Answer:

<h2>2.36 %</h2>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

error = 46.7 - 45.6 = 1.1

actual mass = 46.7

From the question we have

p(\%) =  \frac{1.1}{46.7}  \times 100 \\  = 2.3555...

We have the final answer as

<h3>2.36 %</h3>

Hope this helps you

Alina [70]3 years ago
3 0

Answer:

2.36%

Explanation:

You might be interested in
Suppose 0.245 g of sodium chloride is dissolved in 50. mL of a 18.0 m M aqueous solution of silver nitrate.
Bezzdna [24]

Answer:

\large \boxed{\text{ 0.066 mol/L}}

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Assemble all the data in one place, with molar masses above the formulas and other information below them.

Mᵣ:       58.44  

            NaCl + AgNO₃ ⟶ NaNO₃ + AgCl

m/g:     0.245

V/mL:                 50.

c/mmol·mL⁻¹:       0.0180

2. Calculate the moles of each reactant  

\text{Moles of NaCl} = \text{245 mg NaCl} \times \dfrac{\text{1 mmol NaCl}}{\text{58.44 mg NaCl}} = \text{4.192 mmol NaCl}\\\\\text{ Moles of AgNO}_{3}= \text{50. mL AgNO}_{3} \times \dfrac{\text{0.0180 mmol AgNO}_{3}}{\text{1 mL AgNO}_{3}} = \text{0.900 mmol AgNO}_{3}

3. Identify the limiting reactant  

Calculate the moles of AgCl we can obtain from each reactant.

From NaCl:  

The molar ratio of NaCl to AgCl is 1:1.

\text{Moles of AgCl} = \text{4.192 mmol NaCl} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol NaCl}} = \text{4.192 mmol AgCl}

From AgNO₃:  

The molar ratio of AgNO₃ to AgCl is 1:1.  

\text{Moles of AgCl} = \text{0.900 mmol AgNO}_{3} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol AgNO}_{3}} = \text{0.900 mmol AgCl}

AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.

4. Calculate the moles of excess reactant

                   Ag⁺(aq)  +  Cl⁻(aq) ⟶ AgCl(s)

 I/mmol:      0.900        4.192            0

C/mmol:    -0.900       -0.900        +0.900

E/mmol:      0                3.292          0.900

So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.

5. Calculate the concentration of Cl⁻

\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}

8 0
3 years ago
I need help please help me !
kirill [66]

Answer:

the rust formed because of the water sitting in sied a metal bowl for  long time but they can fix this if they get filters

Explanation:

3 0
3 years ago
What is the boiling point of a solution made by adding 6.69 g of magnesium chloride to 250.0 g of water?. Use the formula of the
olya-2409 [2.1K]

Answer:

100.432ºC

Explanation:

Ebullioscopy is the elevation of the boiling point of a solvent that has a solute nonvolatile dissolved. The change in the temperature may be calculated by Raoult's law:

ΔT = Kb.W.i

Where <em>ΔT</em> is the temperature change, <em>Kb</em> is the ebullioscopy constant,<em> W</em> is the molality and <em>i</em> is the Van't Hoff factor, which determinates the particles that affect the proper.

The molality is:

W = m1/(M1xm2)

Where <em>m1</em> is the solute mass (in g), <em>M1</em> is the molar mass of the solute, and <em>m2</em> is the mass of the solvent (in kg).

The Van't Hoff factor is the number of final particles divided by the number of initial particles. Magnesium is from froup 2, so it forms the cation Mg⁺², and chlorine ins from group 7 and forms the anion Cl⁻, the salt is MgCl₂ and dissociates:

MgCl₂  → Mg⁺²(aq) + 2Cl⁻(aq)

So, it has 1 particle in initial, and 3 in final (1  Mg⁺² and 2 Cl⁻). So:

i = 3/1 = 3.

The molar mass of MgCl₂ is: 24.3 + 2x35.5 = 95.3 g/mol, m1 = 6.69g, m2 = 0.250 kg, so:

W = 6.69/(95.3x0.250)

W = 0.281 m

Then:

ΔT = 0.512x0.281x3

ΔT = 0.432ºC

The normal boling point of water is 100ºC, so

T - 100 = 0.432

T = 100.432ºC

6 0
3 years ago
Refer to the following unbalanced equation:
user100 [1]

Mass of Oxygen (O₂) : = 88.16 g

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

C₆H₁₄+ O₂ → CO₂ + H₂0

25 g C₆H₁₄

Required

mass of oxygen (O₂)

Solution

Balanced equation

2C₆H₁₄ + 19O₂ ⇒12 CO₂ + 14 H₂O

mol C₆H₁₄ (MW=86,18 g/mol) :

= mass : MW

= 25 g : 86.18 g/mol

= 0.29

From the equation, mol ratio of C₆H₁₄ : O₂ = 2 : 19, so mol O₂ :

= 19/2 x mol C₆H₁₄

= 19/2 x 0.29

= 2.755

Mass O₂(MW=32 g/mol) :

= 2.755 x 32

= 88.16 g

6 0
3 years ago
6. The graph below shows the heating curve for ethanol (from –200C to 150C). Calculate the amount of heat (kJ) required for each
Kazeer [188]

This problem is providing the heating curve of ethanol showing relevant data such as the initial and final temperature, melting and boiling points, enthalpies of fusion and vaporization and specific heat of solid, liquid and gaseous ethanol, so that the overall heat is required and found to be 1.758 kJ according to:

<h3>Heating curves:</h3>

In chemistry, we widely use heating curves in order to figure out the required heat to take a substance from a temperature to another. This process may involve sensible heat and latent heat, when increasing or decreasing the temperature and changing the phase, respectively.

Thus, since ethanol starts off solid and end up being a vapor, we will find five types of heat, three of them related to the heating-up of ethanol, firstly solid, next liquid and then vapor, and the other two to its fusion and vaporization as shown below:

Q_T=Q_1+Q_2+Q_3+Q_4+Q_5

Hence, we begin by calculating each heat as follows, considering 1 g of ethanol is equivalent to 0.0217 mol:

Q_1=0.0217mol*111.5\frac{J}{mol*\°C}[(-114.1\°C)-(-200\°C)] *\frac{1kJ}{1000J} =0.208kJ\\&#10;\\&#10;Q_2=0.0217mol*4.9\frac{kJ}{mol} =0.106kJ\\&#10;\\&#10;Q_3=0.0217mol*112.4\frac{J}{mol*\°C}[(78.4\°C)-(-114.1\°C)] *\frac{1kJ}{1000J} =0.470kJ\\&#10;\\&#10;Q_4=0.0217mol*38.6\frac{kJ}{mol} =0.838kJ\\&#10;\\&#10;Q_5=0.0217mol*87.5\frac{J}{mol*\°C}[(150\°C)-(78.4\°C)] *\frac{1kJ}{1000J} =0.136kJ

Finally, we add them up to get the result:

Q_T=0.208kJ+0.106kJ+0.470kJ+0.838kJ+0.136kJ\\&#10;\\&#10;Q_T=1.758kJ

Learn more about heating curves: brainly.com/question/10481356

7 0
2 years ago
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