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2 years ago

Anyone can help me please

Chemistry

1 answer:

Vlad1618 [11]2 years ago

5 0

x=3

Use the pythagorean theorem:

a^2 + b^2 = c^2

4^2 + x^2 = 5^2

16 + x^2 = 25

x^2 = 25-16

x^2 = 9

√x^2 = √9

x = 3

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What is the pH of 0.10 M NaF(aq). The Ka of HF is 6.8 x 10-4

dsp73

Given information:

Concentration of NaF = 0.10 M

Ka of HF = 6.8*10⁻⁴

To determine:

pH of 0.1 M NaF

Explanation:

NaF (aq) ↔ Na+ (aq) + F-(aq)

[Na+] = [F-] = 0.10 M

F- will then react with water in the solution as follows:

F- + H2O ↔ HF + OH-

Kb = [OH-][HF]/[F-]

Kw/Ka = [OH-][HF]/[F-]

At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x

10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x

x = [OH-] = 1.21*10⁻⁶ M

pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92

pH = 14 - pOH = 14-5.92 = 8.08

Ans: (b)

pH of 0.10 M NaF is 8.08

6 0

2 years ago

At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat

wariber [46]

Answer:

0.808 M

Explanation:

Using Raoult's Law

$\frac{P_s}{Pi}= x_i$

where:

$P_s$ = vapor pressure of sea water( solution) = 23.09 mmHg

$P_i$ = vapor pressure of pure water (solute) = 23.76 mmHg

$x_i$ = mole fraction of water

∴

$\frac{23.09}{23.76}= x_i$

$x_i = 0.9718$

$x_i+x_2=1$

$x_2 = 1- x_i$

$x_2 = 1- 0.9718$

$x_2 = 0.0282$

$x_i = \frac{n_i}{n_i+n_2}$ ------ equation (1)

$x_2 = \frac{n_2}{n_i+n_2}$ ------ equation (2)

where; $(n_2) =$ number of moles of sea water

$(n_i) =$ number of moles of pure water

equating above equation 1 and 2; we have :

$\frac{n_2}{n_i}$ $= \frac{0.0282}{0.9178}$

$= 0.02901$

NOW, Molarity = $\frac{moles of sea water}{mass of pure water }*1000$

$= \frac{0.02901}{18}*1000$

$= 0.001616*1000$

$= 1.616 M$

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have $\frac{1.616}{2} =0.808 M$

3 0

2 years ago

Read 2 more answers

What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?

Sergeu [11.5K]

Answer:

$\large \boxed{\text{b. 1.5 h}}$

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

$\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt$

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

$\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}$

2. Calculate the half-life

$t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463 h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}$

4 0

2 years ago

Why do animals need carbohydrates in their diets

Sever21 [200]

Answer:

Carbohydrates are energy-providing feed components composed of carbon, hydrogen, and oxygen. They should make up about 75 percent of an animal's diet. The energy they provide powers muscular movements. Carbohydrates also produce the body heat that helps keep the animal warm.

8 0

2 years ago

What volume of 18.0 m sulfuric acid is required to prepare 16.5 l of 0.126 m h2so4?

Blababa [14]

this is a dilution question where a certain volume is taken from a more concentrated solution and diluted to make a solution with a lower concentration.

the dilution forumla is as follows

c1v1 = c2v2

where c1 is concentration and v1 is volume of the more concetrated solution

c2 is concentration and v2 is volume of the diluted solution

substituting the values in the equation

18.0 M x V = 0.126 M x 16.5 L

V = 0.1155 L

volume of 115.5 mL is taken from the 18.0 M solution and diluted upto 16.5 L solution.

6 0

3 years ago