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Pavel [41]
2 years ago
6

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

evation where the pressure is 0.73 atm, then the new volume is 1.8 L. What is the temperature (in °C) of the air at this elevation?
Chemistry
1 answer:
icang [17]2 years ago
7 0

Answer:

The temperature of the air at this given elevation will be 53.32425°C

Explanation:

We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.

Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm  ; Final Volume = 1.8 L ; Final pressure = 0.73 atm  

We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T_2 ),

P_1V_1 / T_1 = P_2V_2 / T_2,

T_2 = P_2V_2T_1 / P_1V_1,

T_2 = 0.73 atm * 1.8 L * 298.15 K / 1 atm * 1.2 L = ( 0.73 * 1.8 * 298.15 / 1 * 1.2 ) K = 326.47425 K,

T_2 = 326.47425 K = 53.32425 C

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What is the density of an object if its mass is 10 g and a volume of 2 mL?
Lilit [14]

Hi there! Let's solve this problem shall we!

⠀Volume = 10g

      Mass = 2 mL

In this specific problem, they are asking us to find the <u><em>density </em></u>of the object. So,<u><em> using the information given to us</em></u> (volume and mass), let's solve the problem!

Now, if you remember, D = M ÷ V

So, let's fill in the blanks!

D = Our unknown value

M = 2mL

V = 10g

Here is the filled out formula:

D = M ÷ V

D = 2mL ÷ 10g

D = 5 g/mL

*Make sure you put the units for your final solution!*

4 0
1 year ago
A 2.36-gram sample of NaHCO3 was completely decomposed in an experiment.
enyata [817]

Answer:

Explanation:

a) The mass of the reactants is 2.36 grams, and the mass of the products is 1.57 grams plus the mass of the carbonic acid. Thus, using the law of conservation of mass, we get the mass of the carbonic acid is 2.36 - 1.57 =  0.79 grams.

b) The gram-formula mass of sodium bicarbonate is 84.006 g/mol, meaning that 2.36/84.006 = 0.028 moles were consumed. Thus, this means that in theory, 0.014 moles of carbonic acid should have been produced, which would have a mass of (0.014)(62.024)=0.868 grams. Thus, the percentage yield is (0.79)/(0.868) * 100 = 91%

8 0
2 years ago
If you are given the molarity of a solution, what additional information would you need to find the weight/weight percent (w/w%)
Ludmilka [50]

Answer:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

Explanation:

Hello,

In this case, since molarity is mathematically defined as the moles of solute divided by the volume of solution and the weight/weight percent as the mass of solute divided by the mass of solution, we need:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

For instance, if a 1-M solution of HCl has a density of 1.125 g/mL, we can compute the w/w% as follows:

w/w\%=1\frac{molHCl}{L\ sln}*\frac{36.45gHCl}{1molHCl}*\frac{1L\ sln}{1000mL\ sln}*\frac{1mL\ sln}{1.125g\ sln}    *100\%\\\\w/w\%=3.15\%

Whereas the first factor corresponds to the molar mass of HCl, the second one the conversion from L to mL of solution and the third one the density to express in terms of grams of solution.

Regards.

4 0
2 years ago
A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.692 kg of water
S_A_V [24]

Answer:

494.49 g of NaCl.

Explanation:

Data obtained from the question include the following:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mass of NaCl =.?

Next, we shall determine the number of mole of NaCl in the solution.

Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as

Molality = mole of solute /Kg of solvent

With the above formula, we can obtain the number of mole NaCl in the solution as follow:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mole of NaCl =..?

Molality = mole of solute /Kg of solvent

3.140 = mole of NaCl /2.692

Cross multiply

Mole of NaCl = 3.140 x 2.692

Mole of NaCl = 8.45288 moles

Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:

Mole of NaCl = 8.45288 moles

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =.?

Mole = mass /Molar mass

8.45288 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 8.45288 × 58.5

Mass of NaCl = 494.49 g.

Therefore, 494.49 g of NaCl are present in the solution.

8 0
3 years ago
Heating galactose, a monosaccharide sugar, in the presence of excess oxygen produces carbon dioxide gas and water vapor. Classif
artcher [175]

Answer:

it's a combustion reaction

6 0
2 years ago
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