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3 years ago

An object has a mass of 5.003kg and a density of 2.84g/cm^3. Find its volume in cm^3.

Chemistry

1 answer:

pickupchik [31]3 years ago

3 0

Answer:

The answer is

Explanation:

The volume of a substance given the density and mass can be found by using the formula

$volume = \frac{mass}{Density}$

From the question

Density = 2.84 g/cm³

mass = 5.003 kg

We must first convert the mass from kg to grams

If 1 kg = 1000 g

Then 5.003 kg = 5.003 × 1000 = 5003 g

Substitute the values into the above formula and solve

That's

$volume = \frac{5003}{2.84} \\ = 1761.6197$

We have the final answer as

Hope this helps you

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Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 Ka1=6.9×10−3, K a2 = 6.2 × 10 − 8 Ka2=6.2×10−8, and K a3 = 4.8 × 10 −

irina1246 [14]

<u>Answer:</u> To calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

<u>Explanation:</u>

Phosphoric acid is a triprotic acid and it will undergo three dissociation reaction each having their respective dissociation constants.

The chemical equation for the first dissociation reaction follows:

$H_3PO_4\rightleftharpoons H_2PO_4^-+H^+;K_a1=6.9\times 10^{-3}$

The chemical equation for the second dissociation reaction follows:

$H_2PO_4^-\rightleftharpoons HPO_4^{2-}+H^+;K_a2=6.2\times 10^{-8}$

The chemical equation for the third dissociation reaction follows:

$HPO_4^{2-}\rightleftharpoons PO_4^{3-}+H^+;K_a3=4.8\times 10^{-13}$

To form a buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a$ of second dissociation process

To calculate the $pK_a$ , we use the equation:

$pK_a=-\log (K_a)\\\\pK_a=-\log(6.2\times 10^{-8})\\\\pK_a=7.21$

To calculate the pH of buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a2+\log(\frac{[\text{conjugate base}]}{[\text{weak acid}]})$

$pH=pK_a2+\log(\frac{[HPO_4^{2-}]}{[H_2PO_4^-]})$

We are given:

$pK_a2$ = negative logarithm of second acid dissociation constant of phosphoric acid = 7.21

$[HPO_4^{2-}]$ = concentration of conjugate base

$[H_2PO_4^{-}]$ = concentration of weak acid

Hence, to calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

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4 years ago

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Tpy6a [65]

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2 years ago

Cyanogen is a gas which contains 46.2% C and 53.8% N by mass. At a temperature of 25°C and a pressure of 750 mm Hg, 1.50 g of cy

sergiy2304 [10]

Answer:

Molecular formula of cyanogen is C₂N₂

Explanation:

We apply the ideal gases law to find out the mole of cyanogen

P . V = n. R. T

Firstly let's convert the pressure in atm, for R

750 mmHg = 0.986 atm

25°C + 273 = 298K

0.986 atm . 0.714L = n . 0.082 L.atm/mol.K .298K

(0.986 atm . 0.714L) / (0.082 L.atm/mol.K .298K) = n

0.0288 mol = n

Molar mass of cyanogen = mass / mol

1.50 g /0.0288 mol = 52.02 g/m

Let's apply the percent, to know the quantity of atoms

100 g of compound contain 46.2 g of C and 53.8 g of N

52.02 g of compound contain:

(52.02 . 46.2) / 100 = 24 g → 2 atoms of C

(52.02 . 53.8) / 100 = .28 g → 2 atoms of N

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3 years ago