State the name of the ion which is oxidised in the following half equations. Cathode: Na+ + e– → Na Anode: 2Cl– → Cl2 + 2e

What is equal to the number of particles in 24 grams of carbon-12? 2 moles 2:1 mole ratio 2:1 molar mass 1:2 molar volume

mariarad [96]

Answer:

The answer to your question is 2 moles of Carbon

Explanation:

Carbon-12 means it has a molecular mass of 12g / mol. Mol is defined as the number of grams of an atom in one mol.

To solve this problem, we use proportions

12 g of Carbon ---------------- 1 mol of Carbon

24 g of Carbon -------------- x

x = ( 24 g x 1 mol ) / 12 g

x = 24 / 12

x = 2 mol of Carbon

8 0

3 years ago

1. What mass of Mgo will I make from 48 g of Mg?<br> 2Mg + O2 + 2MgO

statuscvo [17]

Answer:

80.6 g

Explanation:

MgO=24.3+16=40.3 g

48.6/24.3= 2 moles

64/2x16= 2 moles

2x40.3= 80.6g

8 0

3 years ago

What is the true ratio of atoms in a molecule or formula unit called?

Colt1911 [192]

Answer:

O the molecular formula

Explanation:

The true ratio of atoms in a molecule or formula unit is called the molecular formula.

The molecular formula of a compound shows the true number of atoms that are combining to form a compound.

For example, the molecular formula of water is H₂O.

Empirical formula of a compound is its simplest ratio.
When the number of atoms are divided by the smallest possible factor, it yields the empirical formula.

3 0

3 years ago

A 0.25-mol sample of a weak acid with an unknown Pka was combined with 10.0-mL of 3.00 M KOH, and the resulting solution was dil

Masteriza [31]

Answer : The value of $pK_a$ of the weak acid is, 4.72

Explanation :

First we have to calculate the moles of KOH.

$\text{Moles of }KOH=\text{Concentration of }KOH\times \text{Volume of solution}$

$\text{Moles of }KOH=3.00M\times 10.0mL=30mmol=0.03mol$

Now we have to calculate the value of $pK_a$ of the weak acid.

The equilibrium chemical reaction is:

$HA+KOH\rightleftharpoons HK+H_2O$

Initial moles 0.25 0.03 0

At eqm. (0.25-0.03) 0.03 0.03

= 0.22

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[HK]}{[HA]}$

Now put all the given values in this expression, we get:

$3.85=pK_a+\log (\frac{0.03}{0.22})$

$pK_a=4.72$

Therefore, the value of $pK_a$ of the weak acid is, 4.72

7 0

3 years ago

If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.

kicyunya [14]

Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

$CH_{3}COOH \ \textless \ ---\ \textgreater \ H^{+} + CH_{3}COO^{-}$

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)

pH = 2.88 ==> $[H^{+}]$ = $10^{-2.88}$ = 0.001 $moldm^{-3}$

The change in Concentration Δ $[CH_{3}COOH]$ = 0.001 $moldm^{-3}$

  CH3COOH H+ CH3COOH
Initial   $moldm^{-3}$      $x$ 0 0

Change $moldm^{-3}$ -0.001 +0.001 +0.001

Equilibrium $moldm^{-3}$ $x$ - 0.001 0.001 0.001


Since the $k_{a}$ value is so small, the assumption
$[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}$ can be made.

$k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}$

Solve for x to get the required concentration.

note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.

2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind

Hope this helps!

8 0

3 years ago

State the name of the ion which is oxidised in the following half equations. Cathode: Na+ + e– → Na Anode: 2Cl– → Cl2 + 2e–