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3 years ago

An atom of carbon contains 6 total electrons. How many of these electrons are in s orbitals?

Chemistry

1 answer:

zaharov [31]3 years ago

6 0

Answer:

Explanation:

total electrons = 6

electronic configuration = 1s^2, 2s^2, 2p^6

so in s orbital there are 4 electrons.

Hope it helps you (｡◕‿◕｡)

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Carry out the following calculation, paying special attention to the significant figures (where 4/3 is exact), rounding, and uni

babunello [35]

Answer:

Value = 1.80 g/cm³ (Approx)

Explanation:

Given:

$\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(1.65 \times 10^2 cm)^3}$

Computation:

$\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(1.65 \times 10^2 cm)^3} \\\\\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(4.492125 \times 10^6 cm^3)} \\\\ \frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(4.492125 \times 10^6 cm^3)}\\\\ \frac{3.39 \times 10^7g}{18.8166132\times 10^6 cm^3} \\\\ 1.80159945g/cm^3$

Value = 1.80 g/cm³ (Approx)

7 0

3 years ago

If 12.5 grams of the original sample of cesium-137 remained after 90.6 years, what was the mass of the original sample?

myrzilka [38]

Answer:

Mass of original sample = 100 g

Explanation:

Half life of cesium-137 = 30.17 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30.17}\ year^{-1}$

The rate constant, k = 0.02297 year⁻¹

Time = 90.6 years

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Initial concentration $[A_0]$ = ?

Final concentration $[A_t]$ = 12.5 grams

Applying in the above equation, we get that:-

$12.5\ g=[A_0]e^{-0.02297\times 90.6}$

$[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g$

Mass of original sample = 100 g

8 0

3 years ago

How many moles of CO gas are in 34.6 L?

Troyanec [42]

moles of CO gas : 1.545

<h3>Further explanation</h3>

Standard conditions for temperature and pressure are used as a reference in certain calculations or conditions

There are 2 conditions that are usually used as a reference : STP and RTP

Assuming the STP state :

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

Then for 34.6 L of CO gas :

$\tt moles=\dfrac{34.6}{22.4}=1.545$

6 0

2 years ago

Which describes a way to speed up the collisions between hydrogen and oxygen molecules to produce more water?

rodikova [14]

Equation:2H2(g) + O2(g) → 2H2O(g)

Smaller container means less volume, and the molecules will hit the walls of the container more frequently because there's less space available and the pressure will go up. I guess this would mean that the side with fewer moles would be favored as a result. We count the number of moles on the reactants and products and find that there are fewer moles on the product side, so I guess this would favor the product formation.

4 0

3 years ago

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All of the following are features of karst topography EXCEPT sinkholes. Disappearing and emerging streams. Caverns. Continuous s

Minchanka [31]

D, continuous streams.

4 0

3 years ago