Which of the following is not a physical property of a liquid?

In a certain city, electricity costs $ 0.18 per kW ⋅ h . What is the annual cost for electricity to power a lamppost for 5.00 h

LuckyWell [14K]

Answer:

The cost of electricity for 100 W power bulb = $ 32.85

Cost of electricity for 0.025 W fluorescent bulb = $ 8.2125

Explanation:

Cost of electricity = $ 0.18 per KW-H

Time = 5 hour per day

Bulb power = 100 W = 0.1 KW

Fluorescent bulb power = 25 W = 0.025 KW

(a) Cost of electricity for 100 W power bulb

0.1 × 5 × 365 × 0.18 = $ 32.85

(b) Cost of electricity for 0.025 W fluorescent bulb

0.025 × 5 × 365 × 0.18 = $ 8.2125

Therefore the cost of electricity for 100 W power bulb = $ 32.85

Cost of electricity for 0.025 W fluorescent bulb = $ 8.2125

6 0

3 years ago

A(g) + 2B(g) → C(g) + D(g)

WITCHER [35]

Answer:

0.169

Explanation:

Let's consider the following reaction.

A(g) + 2B(g) ⇄ C(g) + D(g)

We can find the pressures at equilibrium using an ICE chart.

A(g) + 2 B(g) ⇄ C(g) + D(g)

I 1.00 1.00 0 0

C -x -2x +x +x

E 1.00-x 1.00-2x x x

The pressure at equilibrium of C is 0.211 atm, so x = 0.211.

The pressures at equilibrium are:

pA = 1.00-x = 1.00-0.211 = 0.789 atm

pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm

pC = x = 0.211 atm

pD = x = 0.211 atm

The pressure equilibrium constant (Kp) is:

Kp = pC × pD / pA × pB²

Kp = 0.211 × 0.211 / 0.789 × 0.578²

Kp = 0.169

6 0

3 years ago

What would you need to do to calculate the molality of 10 mol of NaCl in 200

Makovka662 [10]

It will be the second one

8 0

3 years ago

What is the balanced equation and type of reaction for: Copper metal+aqueous silver nitrate

Gnesinka [82]

Answer:

Cu(s) + 2AgNO3(aq)→Cu(NO3)2(aq)+2Ag(s)

This chemical equation means:

One mole of solid copper plus two moles of aqueous silver nitrate produce one mole of copper(II) nitrate plus two moles of solid silver.

This is a single replacement reaction in which the metal copper replaces the metal silver.

3 0

2 years ago

The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3)

cestrela7 [59]

Answer:

a. The limiting reactant is $NaHCO_{3}$

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

a)

You know the following reaction:

$3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}$ ⇒ $3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}$

First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:

Na: 23
H: 1
C: 12
O: 16

To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:

$NaHCO_{3}$ :23+1+12+16*3=84 g/mol
$H_{3} C_{6} HO_{7}$ :1*3+12*6+1*5+16*7= 192 g/mol
$CO_{2}$ :12+16*2= 44 g/mol
$H_{2} O$ :1*2+16= 18 g/mol
$Na_{3} C_{6} H_{5} O_{7}$ : 23*3+12*6+1*5+16*7= 258 g/mol

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of $NaHCO_{3}$ react with 1 mole of $H_{3} C_{6} HO_{7}$ Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 192 g

You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

$grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}$

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So the limiting reagent is sodium bicarbonate.

b)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of $NaHCO_{3}$ react with 3 mole of $CO_{2}$ . Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 132 g

You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

$grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}$

grams of carbon dioxide= 0.73 g

Then, 0.73 g of carbon dioxide are formed.

c)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first. This means that citric acid will not react everything, leaving an excess.

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

$grams of citric acid=\frac{1.4 g * 192 g}{252 g}$

grams of citric acid= 1.067 g

But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

So the grams of excess reactant that do not participate in the reaction are 0333 g.

3 0

3 years ago