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3 years ago

Pls I need help give away 11 points

Chemistry

1 answer:

Maru [420]3 years ago

7 0

Answer:

The charge is +2

Explanation:

Protons are positive, electrons are negative. 10-8=2.

+2 :)

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an open box of bandages is on the floor. bandage wrappers are found nearby. What is the chemical treatment?

katrin2010 [14]

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4 years ago

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The empirical formula of a compound is found to be P2O5. Experiments show that the molar mass of the compound is 426g/mol. What

yulyashka [42]

Answer:

Molecular formula = P₄O₁₀

Explanation:

Given data:

Empirical formula of compound = P₂O₅

Molar mass of compound = 426 g/mol

Molecular formula = ?

Solution:

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass= P₂O₅ = 283.89 g/mol

n = 426 g/mol / 283.89 g/mol

n = 2

Molecular formula = n (empirical formula)

Molecular formula = 2 (P₂O₅ )

Molecular formula = P₄O₁₀

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3 years ago

Suppose a container holds 1000 hydrogen molecules (H2) and 1000 oxygen molecules (O2) that react to form water. How many water m

mezya [45]

500 water molecules and the remaining 500 O2 molecules. Remember the ratio of H to O in H2O.

8 0

4 years ago

A 18.08-g sample of the ionic compound , where is the anion of a weak acid, was dissolved in enough water to make 116.0 mL of so

Oksi-84 [34.3K]

Answer:

a) 129.14 g/mol

b) 8.87

Explanation:

Given that:

mass of the ionic compound [NaA] = 18.08 g

Volume of water = 116.0 mL = 0.116 L

Let the mole of the acid HCl = 0.140 M

Volume of the acid = 500.0 mL = 0.500 L

pH = 4.63

$V_{equivalence}_{acid}$ = 1.00 L

Equation for the reaction can be represented as:

$NaA_{(aq)} + HCl_{(aq)} -----> HA_{(aq)} + NaCl_{(aq)$

From above; 1 mole of an ionic compound reacts with 1 mole of an acid to reach equivalence point = 0.140 M × 1.00 L

= 0.140 mol

Thus, 0.140 mol of HCl neutralize 0.140 mol of ionic compound at equilibrium

Thus, the molar mass of the sample = $\frac{18.08g}{0.140 mole}$

= 129.14 g/mol

b) since pH = pKa

Then pKa of HA = 4.63

Ka = $10^{-4.63]$

= $2.3*10^{-5}$

$[A^-]equ = \frac{0.140M*1.00L}{1.00L+0.116L}$

$= \frac{0.140 mol}{1.116 L}$

= 0.1255 M

$K_a$ of HA = $2.3*10^{-5}$

$K_b = \frac{1.0*10^{-14}}{2.3*10^{-5}}$

$= 4.35*10^{-10}$

$A_{(aq)}$ + $H_2O_{(l)}$ $\rightleftharpoons$ $HA_{(aq)}$ + $OH^-_{(aq)}$

Initial 0.1255 0 0

Change - x + x + x

Equilibrium 0.1255 - x x x

$K_b = \frac{[HA][OH^-]}{[A^-]}$

$4.35*10^{-10} = \frac{[x][x]}{[0.1255-x]}$

As $K_b$ is very small, (o.1255 - x) = 0.1255

$x = \sqrt{0.1255*4.35*10^{-10}}$

[OH⁻] = $x = 7.4 *10^{-6}$

But pOH = - log [OH⁻]

= - log [ $7.4*10^{-6}$ ]

= 5.13

pH = 14.00 = 5.13

pH = 8.87

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3 years ago

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Julli [10]

<span>Chemical reactions at the Earth’s surface </span>

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4 years ago