Lewis structures cannot

Is exploding dynamite a chemical or physical property?

ankoles [38]

<span>the answer would be chemical</span>

5 0

3 years ago

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

A gas is initially at a pressure of 225 kPa and a temperature of 245 K in a container that is 4.5 L. If the gas is compressed to

Artyom0805 [142]

Use the ideal gas equation PV=nRT. You can compare before and after using P1V1/n1T1=P2V2/n2T2. Since the number of moles remains constant you can disregard moles from the equation and use pressure, volume and temp. Make sure your pressure is converted to atmospheres, your volume is in liters, and your temperature is in kelvins.

6 0

3 years ago

How many grams of glucose are needed to prepare 400. g of a 2.00% (m/m) glucose solution g?

Dominik [7]

The grams of glucose are needed to prepare 400g of a 2.00%(m/m) glucose solution g is calculated as below

=% m/m =mass of the solute/mass of the solution x100

let mass of solute be represented by y
mass of solution = 400 g
% (m/m) = 2% = 2/100

grams of glucose is therefore =2/100 = y/400
by cross multiplication

100y = 800
divide both side by 100

y= 8.0 grams

5 0

3 years ago

List the major types of intermolecular forces in order of increasing strength. Is there some overlap? That is, can the strongest

Doss [256]

Answer:

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Explanation:

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6 0

3 years ago