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2 years ago

What geometry does VSEPR predict for the central atom in PF5

1 answer:

8 0

so you can see those fluorine atoms have really spread out around the central phosphorus atom. this gives us a trigonal bi-pyramidal molecular geometry for pf5.

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What would be a simple way to block erosion

Travka [436]

By applying some (compared to other things) simple steps, you can control and prevent soilwearing away! The four most common soil wearing away prevention methods are green plants, geotextiles, mulch, and (big walls to hold back water, soil, etc.). Green plants: The simplest andmost natural way to prevent wearing away is through planting green plants.

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2 years ago

A baseball that is moving through the air has ____________ energy. A. potential B. kinetic

Lunna [17]

Answer:

kinetic

Explanation:

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3 years ago

Question 4 Solve and report to correct number of significant figure 2.375 + 4.5

DiKsa [7]

Answer:

6.9 (two sig figs)

Explanation:

2.375 + 4.5 = 6.875 = 6.9

When adding or subtracting, sig figs are determined by the least number of digits past the decimal point.

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2 years ago

In reality, reactants don't have to react in perfect whole-numbers of moles. In a two-reactant synthesis reaction, usually one

Fofino [41]

Answer:

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2 years ago

Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphe

Gala2k [10]

Answer:

Radius of the interior sphere = 3.847 nm

Explanation:

The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:

$V_s=V_o-V_i=V_i\\V_o=2*V_i\\$

If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):

$r_o=r_i+e$

The first equation becomes

$\frac{4 \pi}{3}*r_o^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\\frac{4 \pi}{3}*(r_i+e)^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\(r_i+e)^{3} = 2 *r_i^{3} \\\\(r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3})=2r_i^{3}\\\\-r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3}=0\\\\-r_i^{3}+3r_i^{2}+3r_i+1=0$

To find ri that satisfies this equation we have to find the roots of the polynomial.

Numerically, it could be calculated that ri=3.847 nm satisfies the equation.

So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.

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2 years ago