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marusya05 [52]
3 years ago
9

The pressure in an automobile tire is 2.0 atm at 27°C. At the end of a journey on a hot summer day the pressure has risen to 2.2

atm. What is the temperature of the air in the tire? a. 272.72 K b. 330 K c. 0.014 K d. 175 K
Chemistry
1 answer:
Kruka [31]3 years ago
3 0

Hey there!

For this we can use the combined gas law:

\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}

We are only working with pressure and temperature so we can remove volume.

\frac{P_{1} }{T_{1}} = \frac{P_{2} }{T_{2}}

P₁ = 2 atm

T₁ = 27 C

P₂ = 2.2 atm

Plug these values in:

\frac{2atm}{27C} = \frac{2.2atm}{T_{2}}

Solve for T₂.

2atm = \frac{2.2atm}{T_{2}}*27C

2atm * T_{2}={2.2atm}*27C

T_{2}={2.2atm}\div2atm*27C

T_{2}=1.1*27C

T_{2}=29.7C

Convert this to kelvin and get 302.85 K, which is closest to B. 330 K.

Hope this helps!

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Covalent compounds are generally not very hard. Justify the statement.
Eduardwww [97]

Covalent compounds are generally not very hard because they are formed by two or more nonmetallic atoms.

<h3>COVALENT COMPOUNDS:</h3>

Covalent compounds are compounds whose constituent elements are joined together by covalent bonds.

Covalent bonding occurs when two or more nonmetallic atoms of an element share valence electrons. This means that covalent compounds will not be physically hard since they constitute non-metals.

Examples of covalent compounds are:

  1. H2 - hydrogen
  2. H2O - water
  3. HCl - hydrogen chloride
  4. CH4 - methane

Learn more about covalent compounds at: brainly.com/question/21505413

4 0
2 years ago
Using stoichiometry, you predict that you should be able to use 314.0 g of Al to produce 1551 g of AlCi3. In your lab
Alex787 [66]

Answer:

90.26%

Explanation:

From the question given above, the following data were obtained:

Theoretical yield of AlCl₃ = 1551 g

Actual yield of AlCl₃ = 1400 g

Percentage yield =?

The percentage yield of the reaction can be obtained as follow:

Percentage yield = Actual yield / Theoretical yield × 100

Percentage yield = 1400 / 1551 × 100

Percentage yield = 140000 / 1551

Percentage yield = 90.26%

Thus, the percentage yield of the reaction is 90.26%

6 0
3 years ago
Sulfuric acid is formed when sulfur dioxide reacts with oxygen and water. Write a balanced chemical equation for the reaction. (
Nesterboy [21]

Balanced chemical equation for the reaction is:

2SO_{2} (g) + O_{2} (g)+ 2H_{2}O (l) ⇒H_{2} S_{} O_{4}

Moles of H_{2} S_{} O_{4} formed is 5.75 moles.

Moles of oxygen used is 5.75 moles in the reaction.

Explanation:

Data given:

moles of SO_{2} = 11.5 moles

moles of H_{2} S_{} O_{4} = ?

Moles of O_{2} needed =?

balanced equation with states of matter =?

Balanced chemical reaction under STP condition is given as:

2SO_{2}(g) + O_{2} (g) + 2H_{2}O (l) ⇒H_{2} S_{} O_{4}

From the balanced reaction 2 moles of sulphur dioxide reacted to form 1 mole of sulphuric acid:

so, from 11.5 moles of SO_{2}, x moles of H_{2} S_{} O_{4} is formed

\frac{1}{2}  =\frac{x}{11.5}

2x = 11.5

x = 5.75 moles of sulphuric acid formed.

From the balanced reaction 1 mole of oxygen reacted to form 1  mole of sulphuric acid.

when 11.5 moles of Sulphur dioxide reacted then oxygen in the reaction is 5.75 moles.

7 0
3 years ago
A 2.000 g sample of CoCl2.xH2O is dried in an oven. When the anhydrous salt is removed from the oven, its mass is 1.565 g. What
Mashutka [201]

The value of X is 2

The total amount of sample taken is 2.00 g

The amount of sample left in the oven after drying is 1.565g

The amount of sample lost (mass of water driven out) = Total sample-Anhydrous salt left in the oven

                                 = 2.00 - 1.565

                                = 0.435 grams

The moles of anhydrous salt present in the hydrate = 1.565g/129.83g/mol = 0.01205

The moles of water present in the hydrate = 0.4350g/18.01g/mol = 0.02415

Therefore the ratio of these two are in 1:2 ratio

The complete chemical reaction is   CoCl2.2H20

To know more about the Similar calculation and explanation click here:

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#SPJ4

6 0
1 year ago
¿Cuáles son las características de los materiales de laboratorio? Por ejemplo: exactitud, resistencia a la temperatura, etc.
boyakko [2]

Answer:

lo siento, no sé punto libre. :p

7 0
3 years ago
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