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Anna11 [10]
3 years ago
15

If you have an atom or copper, an ion of copper and an isotope of copper, what

Chemistry
1 answer:
alexgriva [62]3 years ago
3 0

PLEASE HELP!! PLEASE??

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1. A dry, empty bottle is placed in an ice bath. After the air inside the bottle is cooled off, a balloon is placed
Alex777 [14]

Answer:

hot air rises and expanded the balloon

Explanation:

thats it.

8 0
3 years ago
Read 2 more answers
A particular solution of NaOH has hydronium concentration of 4x10-9 M. what is the solution’s pH?
bearhunter [10]

Answer:

The pH of the solution is 8, 40-

Explanation:

The pH indicates the acidity or basicity of a substance. PH values between 0 and less than 7 indicate acidic solutions, 7 neutral and greater than 7 to 14 basic. It is calculated as

pH = -log (H30+)

pH= -log (4x10-9M)

<em>pH=8,40</em>

8 0
3 years ago
Consider the reaction: 2A(g)+B(g)→3C(g). When A is changing at a rate of -0.110M⋅s−1, How fast is C increasing?
mr_godi [17]
From the reaction above, the rate is given by the following formula:
r = -(1/2) dA / dt = - dB / dt = (1/3) dC/ dt
Note that A and B charge is negative due to they decrease with time
given dA / dt = -0.110 M/s
hence dB / dt = -0.110 / 2 = -0.055 M/s
dC / dt = (-3/2) (-0.110) = 0.165 M/s 
5 0
3 years ago
Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th
Tpy6a [65]

pH = 13.5

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}

The mixture would contain

  • 0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol} of \text{OH}^{-} and
  • 0.1 \times 0.5 = 0.05 \; \text{mol} of \text{Ac}^{-}

if \text{Ac}^{-} undergoes no hydrolysis; the solution is of volume 0.1 + 0.4 = 0.5 \; \text{L} after the mixing. The two species would thus be of concentration 0.30 \; \text{mol} \cdot \text{L}^{-1} and 0.10 \; \text{mol} \cdot \text{L}^{-1}, respectively.

Construct a RICE table for the hydrolysis of \text{Ac}^{-} under a basic aqueous environment (with a negligible hydronium concentration.)

\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}

The question supplied the <em>acid</em> dissociation constant pK_afor acetic acid \text{HAc}; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant pK_b for its conjugate base, \text{Ac}^{-}. The following relationship relates the two quantities:

pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})

... where the water self-ionization constant pK_w \approx 14 under standard conditions. Thus pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3. By the definition of pK_b:

[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b =  10^{-pK_{b}}

x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}

x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}

[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}

pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5

6 0
3 years ago
how many moles of iron will be produced from 6.20 moles of carbon monoxide reacting with excess iron (III) oxide (FeO3) to produ
Gelneren [198K]

0.20 moles of iron will be formed in the reaction.

Explanation:

The balanced chemical equation for the reaction between iron (iii) oxide and carbon monoxide to form Fe is to be known first.

the balanced reaction is :

Fe2O3 + 3CO⇒ 2 Fe + 3 CO2

so from the data given the number of moles of carbon monoxide can be known:

3 moles of CO reacted with Fe2O3 to form 2 moles of iron in the reaction.

Number of moles of CO is 6.20 moles

11.6 gm of iron is formed

so the number of moles of iron formed is calculated as

n = mass of iron ÷ atomic weight of iron

  = 11.6 ÷ 55.84

  = 0.20 moles of iron will be formed when 11.6 gram of iron is produced.

5 0
3 years ago
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