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3 years ago

What is the volume of a salt crystal measuring 1.64 x 10^-2m by 1.5 x 10^-2m by 4.8 x 10^-3

Chemistry

1 answer:

lubasha [3.4K]3 years ago

5 0

Answer:

V = 1.2 × 10⁻⁶ m³

Explanation:

Volume is equal to length × width × height, since we are given these measurements in the question we can simply plug our numbers into the formula.

V = l × w × h

V = 1.64 x 10⁻² × 1.5 x 10⁻² × 4.8 x 10⁻³

V = 1.2 × 10⁻⁶ m³

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What property do all of the group 18 elements have that make them stand out from other elements?

matrenka [14]

They are very stable (not reactive).

This is because their outer shell has a total of 8 valence electrons. All elements strive to get towards a full outer shell, but since these elements already have a full outer shell, they are fairly unreactive.

5 0

2 years ago

If a temperature increase from 21.0 ∘c to 35.0 ∘c triples the rate constant for a reaction, what is the value of the activation

Airida [17]

Answer:

59.077 kJ/mol.

Explanation:

From Arrhenius law: K = Ae(-Ea/RT)

where, K is the rate constant of the reaction.

A is the Arrhenius factor.

Ea is the activation energy.

R is the general gas constant.

T is the temperature.

At different temperatures:

ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]

k₂ = 3k₁ , Ea = ??? J/mol, R = 8.314 J/mol.K, T₁ = 294.0 K, T₂ = 308.0 K.

ln(3k₁/k₁) = (Ea / 8.314 J/mol.K) [(308.0 K - 294.0 K) / (294.0 K x 308.0 K)]

∴ ln(3) = 1.859 x 10⁻⁵ Ea

∴ Ea = ln(3) / (1.859 x 10⁻⁵) = 59.077 kJ/mol.

4 0

2 years ago

4.07 x 10-17 (5.6 x 10") (5.8 x 105)

Andrej [43]

4.07 x 10-17 = 23.7
(5.6 x 10”) (5.8 x 105)= log(100)
log(7)
Hope this helped

3 0

2 years ago

Mixing salt with water is an example of a _________change. * physical change chemical change

SIZIF [17.4K]

Answer:

physical

Explanation:

no chemical reaction is happening

6 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