What did Bohr’s model focus on that Rutherford’s model did not?

A sample of gas in a 14.6 L flexible container is at 25.0oC and 1.00atm. What is the volume of the sample when heated to 220.0oC

inn [45]

Answer: 24.1 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=14.6L\\T_1=25.0^oC=(25+273)K=298K\\V_2=?\\T_2=220.0^0C=(220+273)K=493K$

Putting values in above equation, we get:

$\frac{14.6}{298K}=\frac{V_2}{493}\\\\V_2=24.1L$

Thus the volume of the sample when heated to 220.0oC and the pressure is constant is 24.1 L

7 0

3 years ago

Hi can u help me pls? I'm totally stuck . The natural source of acidity in rain water is _____.

kykrilka [37]

Answer-The correct option is option d with says all of the above.

Explanation- All three acids that are given combined together to form acid rain in which nitric and sulphuric acid are stronger acids present while carbonic acid is a weaker one.

The carbon dioxide admitted in air combines with water to form carbonic acid and gives a weak acidic nature to rainwater. Pollution in nature makes sulphur and nitrogen present in air react to form the stronger acids responsible for acid rain.

5 0

3 years ago

A solution of LiCl in water has XLiCl = 0.0800. What is the molality? A solution of LiCl in water has XLiCl = 0.0800. What is th

Kay [80]

Answer:

mol LiCl = 4.83 m

Explanation:

GIven:

Solution of LiCl in water XLiCl = 0.0800

Mol of water in kg = 55.55 mole

Find:

Molality

Computation:

mole fraction = mol LiCl / (mol water + mol LiCl)

0.0800 = mol LiCl / (55.55 mol + mol LiCl)

0.0800 mol LiCl + 4.444 mol = mol LiCl

mol LiCl - 0.0800 mol LiCl = 4.444 mol

0.92 mol LiCl = 4.444 mol

mol LiCl = 4.83 m

7 0

3 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

What does symbiosis mean in your words. science work

Nat2105 [25]

Answer:

the living together in more or less intimate association or close union of two dissimilar organisms (as in parasitism or commensalism) especially : mutualism.

Explanation:

hope this helps!

8 0

3 years ago