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3 years ago

What is the electrical power generated from a voltage of 12.3 V if the current is 9.42 amps

Chemistry

1 answer:

miv72 [106K]3 years ago

3 0

Answer:

the electrical power would be 115.86 watts. have a nice day

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MC is admitted with pyelonephritis. She has chills, and her temperature is 101 F. She is complaining of flank pain, frequency, a

aleksandr82 [10.1K]

Answer:

i think its B

Explanation:

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3 years ago

What is the correct answer?

Korolek [52]

I think it's Almond Soy Milk because they're recommending your body's pH to be at 7.5 and the Almond Soy Milk is the answer with the closest pH to 7.5

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3 years ago

Give one example of how studying chemistry could be useful in everyday life

Neporo4naja [7]

Answer:

Chemistry is used almost everywhere you go. from the car you use to get to work (gasoline burning) to the nuclear reactors that power many homes ( uranium becoming unstable to product electricity.

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4 years ago

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

4 0

3 years ago

Consider the model of the nitrogen atom.

liq [111]

Answer:

1s2 2s2 2p3

Explanation:

we know that the number of electrons in an atom is equal to number of protons. So the number of electrons here is 7.

Using Moller chart, the electronic configuration is writen by the electrons first enterring into 1s then into 2s after 2p. The s orbital accomodates maximum of 2 electrons.

∴ for atomic no. 7 nitrogen atom, electronic configuration is 1s2 2s2 2p3.

5 0

3 years ago

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