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2 years ago

A shift of a cup of through

Chemistry

2 answers:

Elden [556K]2 years ago

7 0

Huh please be more descriptive, also can y’all answer the questions on my page?

KatRina [158]2 years ago

5 0

What are you asking?

You might be interested in

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

Given that a for HBrO is 2. 8×10^−9 at 25°C. What is the value of b for BrO− at 25°C?

lara [203]

If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

<h3>What is base dissociation constant? </h3>

The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.

The dissociation reaction of hydrogen cyanide can be given as

HCN --- (H+) + (CN-)

Given,

The value of Ka for HCN is 2.8× 10^(-9)

The correlation between base dissociation constant and acid dissociation constant is

Kw = Ka × Kb

Kw = 10^(-14)

Substituting values of Ka and Kw,

Kb = 10^(-14) /{2.8×10^(-9) }

= 3.5× 10^(-6)

Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

DISCLAIMER: The above question have mistake. The correct question is given as

Question:

Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?

learn more about base dissociation constant:

brainly.com/question/9234362

#SPJ4

7 0

1 year ago

Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.

STALIN [3.7K]

Answer: Solution A : $[H_3O^+]=0.300\times 10^{-7}M$

Solution B : $[OH^-]=0.107\times 10^{-5}M$

Solution C : $[OH^-]=0.177\times 10^{-10}M$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

$pH=-\log[H_3O^+]$