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3 years ago

Help please I need it bad

Chemistry

1 answer:

natta225 [31]3 years ago

5 0

= 9.1 × 10^6
(scientific notation)

= 9.1e6
(scientific e notation)

= 9.1 × 10^6
(engineering notation)
(million; prefix mega- (M))

= 9100000
(real number)

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A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical

Katarina [22]

Answer:

empirical formula: $H_2SO_4$

2 g H

32.7 g S

65.3 g O

Explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02 : 1.02/1.02 : 4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is: $H_2SO_4$ .

7 0

3 years ago

450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS

Irina18 [472]

Answer:

600 g K₂SO₄

Explanation:

First write down the complete, balanced chemical equation for such question. In this case:

Cr₂(SO₄)₃ + 2K₃PO₄ → 3K₂SO₄ + 2CrPO₄

Next, calculate molar masses of required compounds mentioned in the question. In this case it is for Cr₂(SO₄)₃ and K₂SO₄.

Molar mass(MM) of Cr₂(SO₄)₃:

= 2*(MM of Cr) + 3*( MM of S) + 3*4*( MM of O)

= 2*(52) + 3*(32) + 12*(16)

= 104 + 96 + 192

= 392 g

Molar mass(MM) of K₂SO₄:

= 2*(MM of K) + 1*( MM of S) + 4*( MM of O)

= 2*(39) + 32 + 4*(16)

= 78 + 32 + 64

= 174 g

Here comes the concept of Limiting reagent:

The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. The other reactants present with this are usually in excess and called excess reactants. If quantities of both the reactants are given, then one should apply unitary method and find out the limiting reagent out of the two. Then, determine the amount of product formed or percentage yield.

Also, 1 mole( 392 g) of Cr₂(SO₄)₃ gives 3 moles( 174*3 = 522 g) of K₂SO₄.

Using unitary method, if 392g of Cr₂(SO₄)₃ gives 522 g of K₂SO₄ , then 450 g of Cr₂(SO₄)₃ will give how much of K₂SO₄?

Yeild of K₂SO₄ : $\frac{522 * 450}{392}$

That is 599.3 g.

Since we have not considered molecular masses of individual atoms to 6 decimal places, this number can be approximated to 600g.

Therefore, 600g of K₂SO₄ is produced.

6 0

3 years ago

Calculate the wavelength, in nanometers, of the spectral line produced when an electron in a hydrogen atom undergoes the transit

Lapatulllka [165]

Answer: The wavelength of spectral line is 656 nm

Explanation:

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.097\times 10^7m^{-1}$

$n_f$ = Final energy level = 2

$n_i$ = Initial energy level = 3

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.524\times 10^6m^{-1}}=6.56\times 10^{-7}m$

Converting this into nanometers, we use the conversion factor:

$1m=10^9nm$

So, $6.56\times 10^{-7}m\times (\frac{10^9nm}{1m})=656nm$

Hence, the wavelength of spectral line is 656 nm

6 0

3 years ago

"Why is Co2, so bad for our bodies and why do we need to remove it?”

zavuch27 [327]

Because it is an asphyxiant.

4 0

2 years ago

Help with the LAB answers

galben [10]

Answer : The reaction rate at $3^oC,24^oC,40^oC\text{ and }65^oC$ are $36mg/L/sec,142mg/L/sec,190mg/L/sec\text{ and }352mg/L/sec$

Solution : Given,

Mass of tablet = 1000 mg

Volume of water = 0.200 L

Formula used :

$\text{Reaction rate}=\frac{\text{Mass of tablet}/\text{Volume of water}}{\text{Reaction time}}$

Now we have to calculate the reaction rate at different temperatures and reaction time.

$\text{Reaction rate at }3^oC=\frac{1000mg/0.200L}{138.5sec}=\frac{5000mg/L}{138.5sec}=36mg/L/sec$

$\text{Reaction rate at }24^oC=\frac{1000mg/0.200L}{34.2sec}=\frac{5000mg/L}{34.2sec}=142mg/L/sec$

$\text{Reaction rate at }40^oC=\frac{1000mg/0.200L}{26.3sec}=\frac{5000mg/L}{26.3sec}=190mg/L/sec$

$\text{Reaction rate at }65^oC=\frac{1000mg/0.200L}{14.2sec}=\frac{5000mg/L}{14.2sec}=352mg/L/sec$

Therefore, the reaction rate at $3^oC,24^oC,40^oC\text{ and }65^oC$ are $36mg/L/sec,142mg/L/sec,190mg/L/sec\text{ and }352mg/L/sec$

5 0

3 years ago

Read 2 more answers