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3 years ago

Are these gases lighter than or denser than air? How can you tell?

1 answer:

3 0

Some Gases are lighter than air
Remember,Air is a mixture of gases
So you cant tell!
It depends on the gas
Hope i helped

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The compound cisplatin, pt(nh3)2cl2 , has been studied extensively as an antitumor agent.

Nataliya [291]

a. Elemental percent composition is the mass percent of each element in the compound.

The formula for mass elemental percent composition = $\frac{mass of element}{mass of compound}$ (1)

The molecular formula of cisplatin is $Pt(NH_3)_2Cl_2$ .

The atomic weight of the elements in cisplatin is:

Platinum, $Pt = 195.084 u$

Nitrogen, $N = 14.0067 u$

Hydrogen, $H = 1 u$

Chlorine, $Cl = 35.453 u$

The molar mass of $Pt(NH_3)_2Cl_2$ = $195.084+ (2\times 14.0067)+(6\times 1)+(2\times 35.453)$ = $300.00 g/mol$

The mass of each element calculated using formula (1):

- Platinum, $Pt$ %

$\frac{195.084}{300.00} \times 100 = 65.23$ %.

- Nitrogen, $N$ %

$\frac{2\times 14.0067}{300.00} \times 100 = 9.34$ %

- Hydrogen, $H$ %

$\frac{6\times 1}{300.00} \times 100 = 2.0$ %

- Chlorine, $Cl$ %

$\frac{2\times 35.453}{300.00} \times 100 = 23.63$ %

b. The given reaction of cisplatin is:

$K_2PtCl_4(aq)+2NH_3(aq)\rightarrow Pt(NH_3)_2Cl_2(s)+2KCl(aq)$

According to the balanced reaction, 1 mole of $K_2PtCl_4$ gives 1 mole of $Pt(NH_3)_2Cl_2$ .

Now, calculating the number of moles of $K_2PtCl_4$ in 100.0 g.

Number of moles = $\frac{given mass}{Molar mass}$

Molar mass of $K_2PtCl_4$ = $2\times 39.0983+195.084+4\times 35.453 = 415.093 g/mol$

Number of moles of $K_2PtCl_4$ = $\frac{100 g}{415.093 g/mol} = 0.241 mole$ .

Since, 1 mole of $K_2PtCl_4$ gives 1 mole of $Pt(NH_3)_2Cl_2$ . Therefore, mass of cisplatin is:

$0.241 mole\times 300 g/mol = 72.3 g$

For mass of $KCl$ :

Molar mass of $KCl$ = $39.0983 + 35.453 = 74.55 g/mol$

Since, 1 mole of $K_2PtCl_4$ gives 2 mole of $KCl$ . Therefore, mass of $KCl$ is:

$0.241 mole\times 74.55 g/mol\times 2 = 35.93 g$

5 0

3 years ago

You wish to measure the iron content of the well water on the new property you are about to buy. You prepare a reference standar

djverab [1.8K]

1.04 ⨯ 10 $^{-4}$ M

<h3>Explanation</h3>

A = ε $\cdot l \cdot c$ by the Beer-Lambert law, where

A the absorbance,
$l$ the path length,
ε the molar absorptivity of the solute, and
$c$ concentration of the solution.

A and ε are the same for both solutions. Therefore, $l \cdot c$ is constant; $l$ is inversely proportional to $c$ . The 100 mL sample would have a concentration 1/4.78 times that of the 45.0 mL reference.

The 13.0 mL standard solution has a concentration of 5.17 ⨯ $10^{-4}$ M. Diluting it to 45.0 mL results in a concentration of $5.17 \times 10^{-4} \times \frac{13.0}{45.0} =$ 1.494 M.

$c$ is inversely related to $l$ for the two solutions. As a result, c₂ = $c_1 \cdot \frac{l_1}{l_2} = 1.494 \times 10^{-4} \times \frac{1}{4.78} =$ 3.126 M.

The 30.0 mL sample has to be diluted by 30.0 / 100.0 times to produce the 100.0 mL solution being tested. The 100.0 mL solution has a concentration of 3.126 M. Therefore, the 30.0 mL solution has a concentration of $3.126 \times \frac{100.0}{30.0} =$ 1.04 ⨯ $10^{-4}$ M.

6 0

3 years ago

A gaseous mixture of o2 and n2 contains 35.8 % nitrogen by mass. what is the partial pressure of oxygen in the mixture if the to

lubasha [3.4K]

The partial pressure is the amount of linguistic compound there is which makes the lagitude of the element 64.663

5 0

3 years ago

The wastewater solution from a factory containing high levels of salts needs to be diluted before it can be released into the en

Vesna [10]

Answer:

3.5 atm

Explanation:

As stated in the question pressure is required to counteract the natural tendency for water to dilute the more concentrated solution. The difference in concentrations will give us the answer using the osmotic pressure equation.

π = ( n/v) RT where n/v is the molarity (mol/L), R is the gas constant and T is the temperature.

The difference in osmotic pressure of the solutions is:

Δπ = Δ c RT where c is the difference in molar concentrations.

pressure required = Δπ = (0.190 - 0.048) M x 0.821 Latm/Kmol x 298 K

= 3.47 atm

3 0

3 years ago

When positively charged particles were radiated onto a gold atom, a few of the particles bounced back. Which of the following ca

MariettaO [177]

Positively charged protons in the nucleus, hope this helps.

8 0

3 years ago

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