I’ll give the brainliest to whoever gets it right!!!

Can you help me understand this ? I completely dont understand it . i know the tire will burst.

icang [17]

Answer:

There is only one formula to use and we should assume ideal gas. This equation is: PV=nRT. For the following questions manipulate this equation to get the answer.

1. n = PV/RT = (249*1000 Pa)(15.6 L)(1 m^3/1000 L)/(8.314 Pa-m^3/mol-K))(21+273) = 1.59 mol

2. P = nRT/V = (1.59)(8.314)(51+273)/(15.6/1000)(1000) = 274.55 kPa

3. Since the answer in #2 is more than 269 kPa, then the tires will likely burst. 4. Reduce pressure way below the limit 269 kPa.

Explanation:

6 0

4 years ago

Read 2 more answers

A 0.56 M solution of AlCl₃ is determined to have a concentration of particles of 1.79 M. What is the van't Hoff factor for AlCl₃

Crank

Answer:

Van't Hoff factor for AlCl₃ = 3 (Approx)

Explanation:

Given:

Number of observed particular = 1.79 M

Number of theoretical particular = 0.56 M

Find:

Van't Hoff factor for AlCl₃

Computation:

Van't Hoff factor for AlCl₃ = Number of observed particular / Number of theoretical particular

Van't Hoff factor for AlCl₃ = 1.79 M / 0.56 M

Van't Hoff factor for AlCl₃ = 3.19

Van't Hoff factor for AlCl₃ = 3 (Approx)

3 0

3 years ago

The total mass of the atmosphere is about 5.00 x 1018 kg. How many moles each of air, O2, and CO2 are present in the atmosphere?

n200080 [17]

<u>Answer:</u> The moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of atmosphere = $5.00\times 10^{18}kg=5.00\times 10^{21}g$

Average molar mass of atmosphere = 28.96 g/mol

Putting values in above equation, we get:

$\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol$

We know that:

Percent of oxygen in air = 21 %

Percent of carbon dioxide in air = 0.0415 %

Moles of oxygen in air = $\frac{21}{100}\times 1.73\times 10^{20}=3.63\times 10^{19}mol$

Moles of carbon dioxide in air = $\frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol$

Hence, the moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

6 0

3 years ago

Some commercially available algaecides for swimming pools claim to contain 7% copper. Could the method used in this experiment t

AleksandrR [38]

Answer:

Explanation:

7% copper implies 7 w/v%(weight/volume %) of copper. This implies a 100 mL algaecide arrangement contains 7 grams of copper.
Henceforth we have a thought of the measure of copper that ought to be available in a given example of algaecide.
Indeed, even a 1 mL algaecide test is sufficient to discover the percentage of copper in it.
so a 25mL sample of algaecide must have around 1.75g of copper.

4 0

3 years ago

Which of the following items is NOT a compound?

Ede4ka [16]

NaOH but you didnt give options

7 0

2 years ago