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dangina [55]
4 years ago
11

Juan works as a tutor for $7 an hour and as a waiter for $14 an hour. This month, he worked a combined total of 103 hours at his

two jobs. Let be the number of hours Juan worked as a tutor this month. Write an expression for the combined total dollar amount he earned this month.
Chemistry
1 answer:
katen-ka-za [31]4 years ago
6 0

Answer:

An expression for the combined total dollar amount he earned this month is $(7x + 14y) where x + y = 103.

Explanation:

Let the no. of hours worked as tutor be x

Earning as tutor in 1 hour = $7

Earning as tutor in x hour = $7*x = $7x

Let the no. of hours worked as waiter be y

Earning as waiter in 1 hour = $14

Earning as waiter in y hour = $14*y = $14y

Given that total hours worked in the month = 103 hours

therefore, x + y = 103 ----> (1)

Total amount earned in the month = Earning as tutor in x hour + Earning as waiter in y hour = $(7x + 14y)

An expression for the combined total dollar amount he earned this month is $(7x + 14y) where x + y = 103.

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A sample of CaCO3(s) is introduced into a sealed container of volume 0.638 L and heated to 1000 K until equilibrium is reached.
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Answer:

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Explanation:

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Now we have to calculate the moles of CO₂

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PV =nRT

P = pressure of gas = 3.8×10⁻²

T = temperature of gas = 1000 K

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\frac{3.8 * 10^2 * 0.638}{0.0821 * 1000} \\= 2.95 * 10^-^4

Now we have to calculate the mass of CaO

mass = 2.95 * 10 ⁻⁴ × 56

= 0.01652g

Therefore,

the mass of CaO present at equilibrium is, 0.01652g

7 0
3 years ago
What is the freezing point of a solution in which 2.50 grams of sodium chloride are added to 230.0 ml of water
labwork [276]

The freezing point of a solution in which 2.5 grams of NaCl is added t0 230 ml of water is :  - 0.69°C

<h3>Determine the freezing point of the solution </h3>

First step : Calculate the molality of NaCl

molality =  ( 2.5 grams / 58.44 g/mol ) / ( 230 * 10⁻³ kg/ml )

              = 0.186  mol/kg

Next step : Calculate freezing point depression temperature

T = 2 * 0.186 * kf

where : kf = 1.86°c.kg/mole

Hence; T = 2 * 0.186 * 1.86 = 0.69°C

Freezing point of the solution

Freezing temperature of solvent - freezing point depression temperature

                                               0°C  -  0.69°C = - 0.69°C

Hence the Freezing temperature of the solution is  - 0.69°C

Learn more about The freezing point of a solution in which 2.5 grams of NaCl is added t0 230 ml of water is :  - 0.69°C

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2 years ago
True or false. an object's weight is constant in all circumstances​
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Ozone (O 3) in the atmosphere can react with nitric oxide (NO): O 3(g) + NO(g) → NO 2(g) + O 2(g). Calculate the ΔG° for this re
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Answer:

ΔG°  = 1022. 8 kJ

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The relationship between these varriables are;

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ΔG°  = 1022. 8 kJ

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