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4 years ago

Choose all dat apply thankssssssssss

Chemistry

1 answer:

slamgirl [31]4 years ago

5 0

Itz defs B and E it might be D also bit iffy on that

YEET GOOD LUCK DADDY

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Given that kb for c6h5nh2 is 1.7 × 10-9 at 25 °c, what is the value of ka for c6h5nh3 at 25 °c?

lina2011 [118]

Answer: $k_a$ for $C_6H_5NH_3^+$ at 25°C is $0.588\times 10^{-4}$

Explanation: We are given $k_b$ of $C_6H_5NH_2$ at 25°C which is $1.7\times 10^{-9}$

To calculate the $k_a$ of $C_6H_5NH_3^+$ , we use the formula:

$k_w=k_a\times k_b$

$k_w\text{ at }25^o=1\times 10^{-14}$

Putting values in above equation, we get:

$1\times 10^{-14}=k_a\times (1.7\times 10^{-9})\\k_a=0.588\times 10^{-5}$

7 0

3 years ago

Given the following UNBALANCED reaction: NH3 (g) <--> N2 (g) + H2 (g) If 1

Yakvenalex [24]

Answer:

C. 1.35

Explanation:

2NH3 (g) <--> N2 (g) + 3H2 (g)

Initial concentration 2.2 mol/0.95L 1.1 mol/0.95L 0

change in concentration 2x x 3x

-0.84 M +0.42M +1.26M

Equilibrium 1.4 mol/0.95L=1.47M 1.58 M 1.26 M

concentration

Change in concentration(NH3) = (2.2-1.4)mol/0.95 L = 0.84M

Equilibrium concentration (N2) = 1.1/0.95 +0.42=1.58 M

Equilibrium concentration(NH3) = 1.4/0.95 = 1.47M

K = [N2]*{H2]/[NH3] = 1.58M*1.26M/1.47M = 1.35 M

8 0

3 years ago

How much pv work is done in kilojoules for the reaction of 0.68 mol of h2 with 0.34 mol of o2 at atmospheric pressure if the vol

dsp73

The delta H of -484 kJ is the heat given off when 2 moles of H2 react with 1 mole of O2 to make 2 moles of H2O. You don't have anywhere near that much reactants, only 1/4 as much

<span>actual delta H = 0.34 moles H2 x (-484 kJ / 2 moles H2) = 823 kJ </span>

<span>delta E = delta H - PdeltaV = 823 kJ - 0.41 kJ = 822 kJ</span>

3 0

4 years ago

Which is most likely water?

svp [43]

Answer:

it looks like a liquid

Explanation:

4 0

3 years ago

Read 2 more answers

Write the net cell equation for this electrochemical cell. Phases are optional. Do not include the concentrations. Co ( s ) ∣ ∣

Tatiana [17]

Answer: The overall equation will be $Co(s)+2Ag^+(aq)\rightarrow Co^{2+}(aq)+Ag(s)$

Explanation:

The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

Anode : $Co(s)\rightarrow Co^{2+}(aq)+2e^{-}$

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

Cathode : $Ag^+(aq)+e^{-1}\rightarrow Ag(s)$ $\times 2$

The number of electrons lost must be equal to the number of electrons gained , thus overall equation will be :

$Co(s)+2Ag^+(aq)\rightarrow Co^{2+}(aq)+Ag(s)$

5 0

3 years ago