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4 years ago

13

The measurement 1.00540 g contains significant figures. * 10 3 4 5 O 6

1 answer:

antiseptic1488 [7]4 years ago

4 0

It contains 6 sig figs

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Why atomic mass is an average vàlue

lukranit [14]

Answer:

The atomic mass is the average number of protons and neutrons for all natural isotopes of an element. It is a decimal number.

Explanation:

Atomic Mass and Mass Number Example :

Hydrogen has three natural isotopes: 1H, 2H, and 3H. Each isotope has a different mass number.

1H has 1 proton. Its mass number is 1. 2H has 1 proton and 1 neutron. Its mass number is 2. 3H has 1 proton and 2 neutrons. Its mass number is 3. 99.98% of all hydrogen is 1H 0.018% of all hydrogen is 2H 0.002% of all hydrogen is 3H Together, they give a value of atomic mass of hydrogen equal to 1.0079 g/mol.

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3 years ago

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HELP ME WITH THIS AND DON'T GET ME WRONG IF YOU DON'T KNOW DON'T ANSWER AND DON'T SAY ''I THINK ITS THE RIGHT ANSWER BUT I'M NOT

tatiyna

Answer:

1 only

Explanation:

a compound is a mix of two different elements, two of the same element is just a molecule

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4 years ago

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What’s inside a cell that allow it to function and survive

mote1985 [20]

Cells have many structure inside of them called organelles which allow it to function and survive

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3 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

6 0

3 years ago

___________ is the rules for speaking or writing a language.

sesenic [268]

Answer:

<u>Grammar</u> is the rules for speaking or writing a language.

3 0

4 years ago