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3 years ago

How much energy is needed to warm 7.40g of water by 55 degress C?

1 answer:

8 0

<span>E = mCdT
E = energy, m = mass, C = specific heat capacity, dT = change in temperature.
526 = 0.074C x 17
E = 0.074C x 55
Divide the equations
E/526 = (0.074C x 55)/(0.074C x 17) = 55/17
E = (55 x 526)/17 = 1702 J</span>

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A salt bridge is used to provide electrons a path to flow from the electrode of one half-cell to another. TRUE FALSE

kicyunya [14]

It is false. The salt bridge is not a path for electrons, but a path for ions to flow from one half-cell to another. It help to balance the charge between the oxidation and reduction vessels.

8 0

3 years ago

How many formula units are there in 50.3 moles potassium chloride?

Alona [7]

<h3>Answer:</h3>

3.03 × 10²⁵ formula units KCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

Reading a Periodic Table
Writing Compounds

<u>Atomic Structure</u>

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

<u>Step 1: Define</u>

50.3 mol KCl (Potassium chloride)

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

<u /> $\displaystyle 50.3 \ mol \ KCl(\frac{6.022 \cdot 10^{23} \ formula \ units \ KCl}{1 \ mol \ KCl} )$ = 3.02907 × 10²⁵ formula units KCl

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.02907 × 10²⁵ formula units KCl ≈ 3.03 × 10²⁵ formula units KCl

6 0

3 years ago

A first-order reaction is 45% complete at the end of 43 minutes. What is the length of the half-life of this reaction

shtirl [24]

The half-life of the reaction is 50 minutes

Data;

Time = 43 minutes
Type of reaction = first order
Amount of Completion = 45%

<h3>Reaction Constant</h3>

Let the initial concentration of the reaction be X $_0$

The reactant left = (1 - 0.45) X $_0$ = 0.55 X $_0$ = X

For a first order reaction

$\ln(\frac{x}{x_o}) = -kt\\ k = \frac{1}{t}\ln (\frac{x_o}{x}) \\ k = \frac{1}{43}\ln (\frac{x_o}{0.55_o})\\ k = 0.013903 min^-^1$

The half-life of a reaction is said to be the time required for the initial amount of the reactant to reach half it's original size.

$x = \frac{x_o}{2} \\t = t_\frac{1}{2} \\t_\frac{1}{2} = \frac{1}{k}\ln(\frac{x_o}{x_o/2})\\$

Substitute the values

$t_\frac{1}{2} = \frac{1}{k}\ln(2)=\frac{0.6931}{0.013903}\\t_\frac{1}{2}= 49.85 min = 50 min$

The half-life of the reaction is 50 minutes

Learn more on half-life of a first order reaction here;

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4 0

2 years ago

2 NH3 + 3 CuO →3 Cu + N₂ + 3 H₂O

Tatiana [17]

40.1g of nitrogen gas is produced.

The equation given is

2 NH₃ + 3 CuO →3 Cu + N₂ + 3 H₂O

This equation is already balanced.

When 3 moles of CuO are consumed, 1 mole of nitrogen gas is produced.

We get 1 mole of nitrogen from 3 moles of copper oxide.

We need to find the number of moles of nitrogen gas produced when 4.3 moles of copper oxide are consumed.

4.3/3 x 1 = 1.433 mols

1.433 mols of nitrogen gas are produced
The molar mass of nitrogen gas is 14+14 = 28g
The amount of nitrogen gas produced in grams is 28x1.433 = 40.1g

40.1g of nitrogen gas can be made when 4.3 moles of CuO are consumed.

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5 0

2 years ago

For every action, there is an equal and opposite reaction.

Svetradugi [14.3K]

Answer:

yeah its newton's third law

Explanation:

7 0

2 years ago