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Alinara [238K]
3 years ago
5

What is the lewis structure for HCL + Na?

Chemistry
1 answer:
saul85 [17]3 years ago
3 0

Answer:

Explanation: that is .....

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Which electron configuration represents an atom of aluminum in an excited state?
weeeeeb [17]
I’m not sure but I think it’s 2-7-4 because the numbers should still add up to 13.(hope it helped)
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3 years ago
Which agent of erosion can create a limestone cave?
Ivanshal [37]
Water containing carbonic acid and calcium
5 0
3 years ago
Given 6.0 mol of N2 are mixed with 12.0 mol of H2. Equation: N2 + 3H2--> 2NH3
andrey2020 [161]
1) Excess reagent

1 mol N2 / 3 mol H2

6.0 mol N2 *3 mol H2 / 1 mol N2 = 18 mol H2

18mol H2 > 12 mol H2 =>  H2 is limiting (you  need 18 mol H2 to use all the 6 mol N2), then N2 is in excees.

12.0 mol H2 * 1mol N2/ 3 mol H2  = 4 mol N2 is the quantity that will react, then the excess is 6 mol N2 - 4 mol N2 = 2 mol N2

2) NH3 produced

12 mol H2 * [2 mol NH3 / 3 mol H2] = 8 mol NH3

Aslso, 4 mol N2 *[2molNH3 / 1 molN2] = 8 mol NH3, the same result.

3) Yield

80% * 8 mol NH3 = 6.4 mol NH3
5 0
3 years ago
An energy of 6.8 x 10^-19 J/atom is required to cause an aluminum atom on a metal surface to lose an electron.
Nana76 [90]

Wavelength of the light is 2.9 × 10⁻⁷ m.

<u>Explanation:</u>

Planck - Einstein equation shows the relationship between the energy of a photon and its frequency, and they are directly proportional to each other and  it is given by the equation as E = hν,

where E is the energy of the photon

h is the Planck's constant = 6.626 × 10⁻³⁴ J s

ν is the frequency

From the above equation, we can find the frequency by rearranging the equation as,

ν = $ \frac{E}{h} = $ \frac{6.8 \times 10^{-19}}{6.626\times10^{-34}} = 1.03\times10^{15} s^{-1}

Now the frequency and the wavelength are in inverse relationship with each other.

ν × λ = c

It can be rearranged to get λ as,

λ = c / ν

 = $\frac{3\times 10^{8} ms^{-1}}{1.03\times10^{15}s^{-1}} = 2.9\times 10^{-7} m

So wavelength is 2.9 × 10⁻⁷ m.

6 0
4 years ago
Doing an experiment about electroplating, you attempt to coat silver in gold using a basic electroplating set-up. You take the m
Elan Coil [88]

Answer:

it’s mass was greater than when it started

Explanation:

When a metal is coated with another metal, the plating metal deposits on the plated metal. Usually, the plating metal functions as the anode while the plated metal functions on the cathode. The anode metal is oxidized and reduced at the cathode and become deposited on the cathode material. This increases the mass of the cathode. Hence the mass of the silver/gold product is greater than the mass of silver at the beginning of the electroplating process.

3 0
3 years ago
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