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Anna71 [15]
2 years ago
7

Give the full electron configuration for nitrogen.

Chemistry
1 answer:
garri49 [273]2 years ago
3 0

Answer:

1s2 2s2 2p3

Explanation:

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A gas with an empirical formula C2H2O has a molecular weight of 120.6g/mol. A possible molecular formula for the gas is
Virty [35]

M(C2H2O)= 12.0*2 +1.0*2 +16.0 = 42 g/mol is a molar mass for empirical formula.

120.6g/mol/42g/mol ≈ 3

So, empirical formula should be increased 3 times,

and molecular formula is C6H6O3.

Answer is D.

7 0
3 years ago
For an enzyme that displays Michaelis-Menten kinetics, what is the reaction velocity, v(as a percentage of Vmax), observed at:a)
fiasKO [112]

Answer:

a) 50% of the maximum velocity

b) 33.33% of the maximum velocity

c) 9.09% of the maximum velocity

d) 66.66% of the maximum velocity

e) 90.9% of the maximum velocity

Explanation:

The Michaelis-Menten kinetis is represented by

v = Vmax*S/(Km+S)

where

v= reaction rate

S= Substrate's concentration

Vmax= maximum rate of reaction

Km= constant

a) for S=Km

v = Vmax*Km/(2Km) = Vmax/2

v/Vmax = 1/2= 50% of the maximum velocity

b) for S=Km/2

v = Vmax*(Km/2)/(3/2Km) = Vmax/3

v/Vmax = 1/3= 33.33% of the maximum velocity

c) for S= 0.1*Km=Km/10

v = Vmax*(Km/10)/(11/10Km) = Vmax/11

v/Vmax = 1/11= 9.09% of the maximum velocity

d) for S=2*Km

v = Vmax*(2*Km)/(3*Km) = (2/3)* Vmax

v/Vmax = 2/3 = 66.66% of the maximum velocity

d) for S=10*Km

v = Vmax*(10*Km)/(11*Km) = (10/11)* Vmax

v/Vmax = 10/11 = 90.9 % of the maximum velocity

7 0
3 years ago
1. Number the following steps of the quantum leap process in order from beginning to end.
kari74 [83]

Answer:

1. Atom absorbs energy from a source

2. Electron becomes unstable because it is quantized

3. Electron jumps to a higher energy level

4. Energy is released in the form of light

5. Electron falls to a lower energy level

5 0
2 years ago
If 45.00 g of precipitate is formed from the reaction of 0.100 mol/L
Tom [10]

Answer:

Approximately 2.53\; \rm L (rounded to three significant figures) assuming that {\rm HCl}\, (aq) is in excess.

Explanation:

When {\rm HCl} \, (aq) and {\rm AgNO_3}\, (aq) precipitate, {\rm AgCl} \, (s) (the said precipitate) and \rm HNO_3\, (aq) are produced:

{\rm HCl}\, (aq) + {\rm AgNO_3}\, (aq) \to {\rm AgCl}\, (s) + {\rm HNO_3}\, (aq) (verify that this equation is indeed balanced.)

Look up the relative atomic mass of \rm Ag and \rm Cl on a modern periodic table:

  • \rm Ag: 107.868.
  • \rm Cl: 35.45.

Calculate the formula mass of the precipitate, \rm AgCl:

\begin{aligned}& M({\rm AgCl})\\ &= (107.868 + 35.45)\; \rm g \cdot mol^{-1} \\\ &\approx 143.318 \; \rm  g\cdot mol^{-1}\end{aligned}.

Calculate the number of moles of \rm AgCl formula units in 45.00\; \rm g of this compound:

\begin{aligned}n({\rm AgCl}) &= \frac{m({\rm AgCl})}{M({\rm AgCl})} \\ &\approx \frac{45.00\; \rm g}{143.318\; \rm g \cdot mol^{-1}}\approx 0.313987\; \rm mol \end{aligned}.

Notice that in the balanced equation for this reaction, the coefficients of {\rm AgNO_3} \, (aq) and {\rm AgCl}\, (s) are both one.

In other words, if {\rm HCl}\, (aq) (the other reactant) is in excess, it would take exactly 1\; \rm mol of {\rm AgNO_3} \, (aq)\! formula units to produce 1\; \rm mol \! of {\rm AgCl}\, (s)\! formula units.

Hence, it would take 0.313987\; \rm mol of {\rm AgNO_3} \, (aq)\! formula units to produce 0.313987\; \rm mol\! of {\rm AgCl}\, (s)\! formula units.

Calculate the volume of the {\rm AgNO_3} \, (aq)\! solution given that the concentration of the solution is 0.124\; \rm mol \cdot L^{-1}:

\begin{aligned}V({\rm AgNO_3}) &= \frac{n({\rm AgNO_3})}{c({\rm AgNO_3})} \\ &\approx \frac{0.313987\; \rm mol}{0.124\; \rm mol \cdot L^{-1}}\approx 2.53\; \rm L\end{aligned}.

(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of {\rm AgNO_3} \, (aq)\!.)

In other words, approximately 2.53\; \rm L of that {\rm AgNO_3} \, (aq)\! solution would be required.

4 0
2 years ago
10. A toy car travels 100 cm in 20 seconds traveling west. What is the velocity of the toy car? O 5cm/s 50 cm/s 5 cm/s west 50 c
astraxan [27]

5cm

Explanation:

you have to divide 100cm by 20 sec

7 0
3 years ago
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