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Kaylis [27]
3 years ago
10

PF3Br2 is a nonpolar molecule. Based on this information, determine the F−P−F bond angle, the Br−P−Br bond angle, and the F−P−Br

bond angle. Enter the number of degrees of the F−P−F, Br−P−Br, and F−P−Br bond angles, separated by commas (e.g., 30,45,90)
Chemistry
1 answer:
inna [77]3 years ago
8 0

Answer:

120,180,90.

Explanation:

PF3Br2 has a trigonal bypiramidal geometry. The three atoms of F will be arranged around the P atom in the same plane and each F-P-F bond will have an angle of 120 degrees. There will be only one Br atom above and beneath the P atom so the bond angle for Br-P-Br will be 180 degrees. Finally, the F−P−Br bond angle will be at 90 degrees.

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How do you know a physical change has occurred when evaporating water produces gaseous water?
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Answer:

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6 0
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What is the molarity of a NaCl stock used to make 750 ml of a dilute 10 mM solution if 5 ml of the concentrated solution was use
ch4aika [34]

Explanation:

The given data is as follows.

          M_{1} = 10 mM = 10 \times 10^{-3} M

          V_{1} = 750 ml,           V_{2} = 5 ml

          M_{2} = ?

Therefore, calculate the molarity of given NaCl stock as follows.

                  M_{1} \times V_{1} = M_{2} \times V_{2}

                  10 \times 10^{-3} \times 750 ml = M_{2} \times 5 ml

                   M_{2} = 1.5 M

Thus, we can conclude that molarity of given NaCl stock is 1.5 M.        

7 0
3 years ago
Please someone help mee
Nikitich [7]

Answer:

13.9kj/mol

Explanation:

h = 891 -(393.5 +2*241.8)=1768.1=13.9kj/mol

8 0
3 years ago
Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo
Romashka-Z-Leto [24]

Answer:

\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

Explanation:

Volume of a cone:

  • \displaystyle V=\frac{1}{3} \pi r^2 h

We have \displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec} and we want to find \displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ? when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

  • \displaystyle V =\frac{1}{3} \pi (5h)^2 h  
  • \displaystyle V =\frac{1}{3} \pi \ 25h^3

Differentiate this equation with respect to time t.

  • \displaystyle \frac{dV}{dt}  =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}
  • \displaystyle \frac{dV}{dt}  =25 \pi h^2 \ \frac{dh}{dt}

Plug known values into the equation and solve for dh/dt.

  • \displaystyle 10 = 25 \pi (2)^2  \ \frac{dh}{dt}
  • \displaystyle 10 = 100 \pi  \ \frac{dh}{dt}  

Divide both sides by 100π to solve for dh/dt.

  • \displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}
  • \displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

The height of the cone is increasing at a rate of 1/10π cm per second.

7 0
3 years ago
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