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2 years ago

List three factors that may affect the rate at which the metal reacts with a given acid

Chemistry

1 answer:

Novay_Z [31]2 years ago

8 0

Reactant concentration, the physical state of the reactants, and surface area, temperature.

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Base your answers on the graph below, which represents uniform cooling of a sample of a pure substance, starting as a gas. Solid

Karolina [17]

Answer:

D & E

Explanation:

I think this is dealing with latent heat and D & E would be the range where you will find solid and liquid phases in equilibrium, cuz it starts as gas at from A to B, B to C is gas and liquid equilibrium, C to D is liquid, D to E solid and liquid, and then E to F is solid.

7 0

2 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

Using the following equation for the combustion of octane calculate the heat associated with the formation of 100.0 g of carbon

natali 33 [55]

Answer:

The right solution is "-602.69 KJ heat".

Explanation:

According to the question,

The 100.0 g of carbon dioxide:

= $\frac{100.0 \ g}{114.33\ g/mole}$

= $0.8747 \ moles$

We know that 16 moles of $CO_2$ formation associates with -11018 kJ of heat, then

0.8747 moles $CO_2$ formation associates with,

= $-\frac{0.8747}{16}\times 11018 \ KJ \ of \ heat$

= $-0.0547\times 11018$

= $-602.69 \ KJ \ heat$

8 0

2 years ago

What is the stoichiometric ratio between BaCl2 and NaCl

bixtya [17]

<span>BaCl2+Na2SO4---->BaSO4+2NaCl There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent? "First convert grams into moles" 1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2 "From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl" The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>

7 0

3 years ago

A form of energy produced my a machine

sergij07 [2.7K]

Sound, heat, light etc

3 0

2 years ago