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3 years ago

Please help! ( number 2)

Chemistry

2 answers:

Fittoniya [83]3 years ago

6 0

They are all metamorphic but gneiss can be <span>formed by the metamorphosis of granite, or sedimentary </span><span>rock. </span>

GuDViN [60]3 years ago

3 0

All of them is a type of a metamorphic rock. I searched it up and it showed all of the answers so tell your teacher all of them is a type of metamorphic rock.

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The conversion of one mole of water from the solid to the liquid phase at the phase transition temperature requires the absorpti

dybincka [34]

Answer: heat

Explanation:

Water at solid is called ICE. This ice absorbs HEAT from its environment, and undergoes a process called melting to convert to liquid, this transition occurs at the melting point of ice

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What are substances called whose solutions conduct electricity?

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It is called an electrolyte

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Why are chemical sanitizers ineffective in killing microbes present in cracks and crevices?

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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

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3 years ago

Name at least two Benefits of using models in science

Alex787 [66]

They provide a way to understand the processes that might otherwise be outside the scope of humans. Also they provide scientists with foundations for further experiments and hypothesis.

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