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Alex787 [66]
3 years ago
13

Anyone in connections taking chemistry that can help me? P.S this IS a question so do not delete it.

Chemistry
1 answer:
iren [92.7K]3 years ago
6 0
I suggest watching Martin Shkerli YouTube channel. He has a ton of videos on chemistry and is very helpful. 
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What reaction type is depicted by the following equation? 2C2H2 + 5O2 4CO2 + 2H2O decomposition combination combustion single-re
Vladimir [108]

combustion is when you burn something in presence e of oxygen and the products will be mainly the CO2 and H2O

3 0
3 years ago
Read 2 more answers
A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after
choli [55]

Answer:

9.25

Explanation:

Let first find the moles of NH_4Cl and NaOH

number of moles of NH_4Cl = 0.40  mol/L × 200 ×  10⁻³L

= 0.08 mole

number of moles of NaOH = 0.80  mol/L × 50 ×  10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

The ICE Table is shown below as follows:

                            NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

Initial (M)              0.08            0.04                            0

Change (M)         - 0.04          -0.04                          + 0.04

Equilibrium (M)      0.04             0                                0.04

K_a*K_b = 10^{-14} \ at \ 25^0C

K_a = \frac{10^{-14}}{K_b}

K_a = \frac{10^{-14}}{1.76*10^{-5}}

K_a= 5.68*10^{10}

pK_a = - log \ (K_a)

pK_a = - log \ (5.68*10^{-10})

pK_a = 9.25

pH = pKa + \ log (\frac{HB}{HA} )   for buffer solutions

pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} ) since they are in the same solution

pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )

pH = 9.25

8 0
3 years ago
PLZ HELP
nika2105 [10]
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8 0
3 years ago
Read 2 more answers
Which of the following is/are true regarding initial and maximal velocity of enzyme-mediated reactions?
kenny6666 [7]

Answer:

a. True

b. True

c. False

d. True

Explanation:

a). A a very low substrate concentration , $S. Thus according to the Machaelis-Menten equation becomes

$V_0 = \frac{V_{max} \times [S]}{Km}$

Here since the $V_0$ varies directly to the substrate concentration [S], the initial velocity is lower than the maximal velocity. Thus option  (a) is true.

b). The Michaelis -Menten kinetics equation states that :

   $V_0 = \frac{V_{max} \times [S]}{Km+[S]}$

  Here the initial velocity changes directly with the substrate concentration as $V_0$ is directly proportional to [S]. But $V_{max}$ is same for any particular concentration of the enzymes. Thus, option (b) is true.

c). As the substrate concentration increases, the initial velocity also increases. Thus option (c) is false.

d). Option (d) explains the procedures to estimate the initial velocity which is correct. Thus, option (d) is true.

7 0
3 years ago
An explanation for a broad of observation, facts, and tested hypothses is called a
nikitadnepr [17]

Answer:

that's is a theory

Explanation:

hope it may help you

4 0
2 years ago
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