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3 years ago

Percent Error Problems – Find the percent error for each problem below.

Chemistry

1 answer:

Citrus2011 [14]3 years ago

5 0

% error = 3.4 %

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

∴ % error = |355 mL – 343 mL|/355 mL × 100 % = |12|/355 × 100 % = 3.4 %

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PCl3(g) + Cl2(g) ⇋ PCl5(g) Kc = 91.0 at 400 K. What is the [Cl2] at equilibrium if the initial concentrations were 0.24 M for PC

Dmitry_Shevchenko [17]

Answer:

[Cl₂] in equilibrium is 1.26 M

Explanation:

This is the equilibrium:

PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)

Kc = 91

So let's analyse, all the process:

PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)

Initially 0.24 M 1.50M 0.12 M

React x x x

Some amount of compound has reacted during the process.

In equilibrium we have

0.24 - x 1.50 - x 0.12 + x

As initially we have moles of product, in equilibrium we have to sum them.

Let's make the expression for Kc

Kc = [PCl₅] / [Cl₂] . [PCl₃]

91 = (0.12 + x) / (0.24 - x) ( 1.50 - x)

91 = (0.12 + x) / (0.36 - 0.24x - 1.5x + x²)

91 (0.36 - 0.24x - 1.5x + x²) = (0.12 + x)

32.76 - 158.34x + 91x² = 0.12 +x

32.64 - 159.34x + 91x² = 0

This a quadratic function:

a = 91; b= -159.34; c = 32.64

(-b +- √(b² - 4ac)) / 2a

Solution 1 = 1.5

Solution 2 = 0.23 (This is our value)

So [Cl₂] in equilibrium is 1.50 - 0.23 = 1.26 M

5 0

3 years ago

A 110 g copper bowl contains 240 g of water, both at 21.0°C. A very hot 410 g copper cylinder is dropped into the water, causing

vlada-n [284]

Answer:

There is 98.76 kJ energy transfered to the water as heat.

Explanation:

Step 1: Data given

Mass of copper bowl = 110 grams

Mass of water = 240 grams

Temperature of water and copper = 21.0 °C

Mass of the hot copper cylinder = 410 grams

8.6 grams being converted to steam

Final temperature = 100 °C

Step 2: Calculate the energy gained by the water:

Q = m(water)*C(water)*ΔT + m(vapor)*Lw

⇒with mass of water = 0.240 kg

⇒ with C(water) = the heat capacity of water = 4184 J/kg°C

⇒ with ΔT = the change in temperature = T2 - T1 = 100 °C - 21.0 = 79°C

⇒ with mass of vapor = 8.60 grams = 0.0086 kg

⇒ with Lw = The latent heat of vaporization (water to steam) = 22.6 *10^5 J/kg

Q = 0.24kg * 4184 J/kg°C *79°C + 0.0086 kg*22.6*10^5 J/kg

Q = 79328.64 + 19436 = 98764.64 J = 98.76 kJ

There is 98.76 kJ energy transfered to the water as heat.

4 0

3 years ago

Calculate the number of liters in 3.25 g of ammonia

il63 [147K]

The liters in 3.25 g of ammonia 4.28 L

calculation

Step 1: find moles of ammonia

moles = mass÷ molar mass

From periodic table the molar mass of ammonia (NH₃) = 14 +(1×3 ) = 17 g/mol

3.25 g÷ 17 g/mol = 0.191 moles

Step 2: find the number of liters of ammonia

that is at STP 1 moles = 22.4 L

0.191 moles = ? L

by cross multiplication

={( 0.191 moles ×22.4 L) / 1 mole} = 4.28 L

8 0

3 years ago

Read 2 more answers

The second atomic structure YES No

gizmo_the_mogwai [7]

Answer

The second part of the theory says all atoms of a given element are identical in mass and properties. The third part says compounds are combinations of two or more different types of atoms. The fourth part of the theory states that a chemical reaction is a rearrangement of atoms.

Explanation:

4 0

2 years ago

A scientist is raising the temperature of a

Arlecino [84]

Answer:

nothing

Explanation:

Here no result have been found yet..

Therfore we cannot determine which variable is dependent and which is not

8 0

1 year ago