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3 years ago

Sulfuric acid dissolves aluminum metal according to the following reaction:

Chemistry

1 answer:

Fiesta28 [93]3 years ago

3 0

Answer:

$m_{H_2SO_4}=81.7gH_2SO_4$

$m_{H_2}=1.67gH_2$

Explanation:

Hello,

Based on the given undergoing chemical reaction is is rewritten below:

$2Al (s) + 3H_2SO_4 (aq)\rightarrow Al _2(SO4)_3 (aq) + 3H_2 (g)$

By stoichiometry we find the minimum mass of H2SO4 (in g) as shown below:

$m_{H_2SO_4}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2SO_4}{2molAl}*\frac{98gH_2SO_4}{1molH_2SO_4} \\m_{H_2SO_4}=81.7gH_2SO_4$

Moreover, mass of H2 gas (in g) would be produced by the complete reaction of the aluminum block turns out:

$m_{H_2}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2}{2molAl}*\frac{2gH_2}{1molH_2} \\m_{H_2}=1.67gH_2$

Best regards.

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How does temperature affect winds all over the globe?

olga2289 [7]

Answer:

The differences in atmospheric pressure generate winds.

Explanation:

At the Equator, the sun warms the water and land more than it does the rest of the world. Warm equatorial air rises higher into the atmospher and migrates towards the poles. The complex relationships between fronts cause different types of winds and weather patterns.

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2 years ago

Which of the following is a nonrenewable energy source?

Ipatiy [6.2K]

B fossil energy

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3 years ago

Polonium- 210 , Po210 , decays to lead- 206 , Pb206 , by alpha emission according to the equation Po84210⟶Pb82206+He24 If the ha

elixir [45]

Answer:

0.269 g

Explanation:

$Po_{210} ^{84}$ ⟶ $Pb_{206} ^{82}$ + $+ He_{4} ^{2}$

Data:

Half-life of $Po_{210} ^{84}$ (T(1/2)) = 138.4 days

Mass of PoCl4 = 561 mg (0,561 g) and molecular weight of PoCl4 = 350. 79 g/mol

Time = 338.8 days

Isotopic masses

$Po_{210} ^{84}$ = 209.98 g/mol

$Pb_{206} ^{82}$ = 205.97 g/mol

Concepts

Avogadro’s number: This is the number of constituent particles that are contained in a mol of any substance. These constituted particles can be atoms, molecules or ions). Its value is 6.023*10^23.

The radioactive decay law is

N=Noe^(-λt)

Where:

No = number of atoms in t=0

N = number of atoms in t=t (now) in this case t=338.8 days

λ= radioactive decay constant

The radioactive constant is related to the half-life by the next equation

λ= $\frac{ln 2}{t(1/2)}$

λ= $\frac{ln2}{138.4 days}$ =0,005008 days^(-1)

No (Atoms of $Po_{210} ^{84}$ in t=0)

To get No we need to calculate the number of atoms of $Po_{210} ^{84}$ in the initial sample. We have a sample of 0,561 g of PoCl4. If we get the number of moles of PoCl4 in the sample, this will be the number of moles of $Po_{210} ^{84}$ in the initial sample.

This is:

$\frac{0,561 g of PoCl4}{350. 79 g of PoCl4 /mol}$ = 0,001599 mol of PoCl4

This is the number of mol of $Po_{210} ^{84}$ in the initial sample.

To get the number of atoms in the initial sample we use the Avogadro’s number = 6.023*10^23

0,001599 mol of $Po_{210} ^{84}$ * 6.023*10^23 atoms/ mol of $Po_{210} ^{84}$ = 9.632 *10^20 atoms of $Po_{210} ^{84}$

Atoms after 338.8 days

We use the radioactive decay law to get this value

N=Noe^(-λt)

N=9.632*10^20 e^(-0,005008 days^(-1) * 338.8 days) =1.765*10^20

This is the number of atoms of $Po_{210} ^{84}$ in the sample after 338.8 days has passed

The number of atoms $Po_{210} ^{84}$ transformed is equal to the number of atoms of $Pb_{206} ^{82}$ produced.

The number of atoms of $Po_{210} ^{84}$ transformed is No - N

9.632 *10^20 – 1.765 *10^20 = 7.866*10^20

So, 7.866*10^20 is the number of atoms of $Pb_{206} ^{82}$ produced

We can get the mass with the Avogadro’s number

(7.866*10^20 atoms of $Pb_{206} ^{82}$ ) / ( 6.023*10^23 atoms of $Pb_{206} ^{82}$ / mol of $Pb_{206} ^{82}$ = 0.001306 moles of $Pb_{206} ^{82}$

This number of moles have a mass of:

(0,001306 moles of $Pb_{206} ^{82}$ )* (205.97 g of $Pb_{206} ^{82}$ /mol of $Pb_{206} ^{82}$ ) = 0.269 g

3 0

3 years ago

a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re

Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Given : 59.4 g of $H_2SO_4$ in 100 g of solution

moles of $H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61$

Volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.61\times 1000}{54.6ml}=11.2M$

To calculate the volume of acid, we use the equation given by neutralisation reaction:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock acid which is $H_2SO_4$

$M_2\text{ and }V_2$ are the molarity and volume of dilute acid which is $H_2SO_4$

We are given:

$M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL$

Putting values in above equation, we get:

$11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL$

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0

3 years ago

A sample of gas contains 6.25 × 10-3 mol in a 500.0 mL flask at 265°C. What is the pressure of the gas in kilopascals? Which var

RideAnS [48]

55.9 kPa; Variables given = volume (V), moles (n), temperature (T)

We must calculate p from V, n, and T, so we use the Ideal Gas Law:

pV = nRT

Solve for p: p = nRT/V

R = 8.314 kPa.L.K^(-1).mol^(-1)

T = (265 + 273.15) K = 538.15 K

V = 500.0 mL = 0.5000 L

∴ p = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa

4 0

3 years ago

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