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3 years ago

Atomic Number: 10 Is: 2s: 2p: 35: 3p: 4s: 3d: 4p: 5s:

Chemistry

1 answer:

aivan3 [116]3 years ago

4 0

Answer:

Atomic Number: 10

Is: Neon

2s: Helium

2p: Helium

35: Bromine

3p: argon

4s: Beryllium

3d: Not sure

4p: Beryllium

5s: Boron

Hope this helped :)

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Which of these are a bronsted acid, a bronsted base or are both the bronsted acid and the bronsted base?

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We can classify the ion molecules as follows:

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For a Bronsted-Lowry acid and bases, an acid is a proton (hydrogen ion) donor. A base is a proton (hydrogen ion) acceptor. Hope this answers the question. Have a nice day.

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3 years ago

If an atom contains more electrons than protons, it is

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Explanation:

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3 years ago

Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is S2(g)+C(s)↽−−⇀CS2(g)????c=9.40 at 900 K Ho

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Answer : The mass of $CS_2$ is, 555.028 grams

Explanation :

First er have to calculate the concentration of $S_2$ .

$\text{Concentration of }S_2=\frac{\text{Moles of }S_2}{\text{Volume of solution}}=\frac{8.08mole}{5.35L}=1.51mole/L$

Now we have to calculate the concentration of $CS_2$ .

The given balanced chemical reaction is,

$S_2(g)+C(s)\rightleftharpoons CS_2(g)$

Initial conc. 1.51 0 0

At eqm. conc. (1.51-x) (x) (x)

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CS_2]}{[S_2]}$

Now put all the given values in this expression, we get :

$9.40=\frac{x}{(1.51-x)}$

By solving the term 'x', we get :

x = 1.365 M

Concentration of $CS_2$ = x M = 1.365 M

Now we have to calculate the moles of $CS_2$ .

$\text{Moles of }CS_2=\text{Concentration of }CS_2}\times \text{Volume of solution}=1.365mole/L\times 5.35L=7.303mole$

Now we have to calculate the mass of $CS_2$ .

Molar mass of $CS_2$ = 76 g/mole

$\text{Mass of }CS_2=\text{Moles of }CS_2}\times \text{molar mass of }CS_2}=7.303mole\times 76g/mole=555.028g$

Therefore, the mass of $CS_2$ is, 555.028 grams

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4 years ago

Carbon-14 has a half-life of 5730 years. In a plant fossil, you find that the 14C has decayed to 1/8.00 of the original amount.

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Take three half lives.

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4 years ago

An industrial synthesis of urea obtains 87.5 kg of urea upon reaction of 68.2 kg of ammonia with excess carbon dioxide. Determin

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Answer:

The theoretical yield of urea = 120.35kg

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Explanation:

Lets calculate -

The given reaction is -

$2NH_3(aq)+CO_2$ → $CH_4N_2O(aq)+H_2O (l)$

Molar mass of urea $CH_4N_2O$ = 60g/mole

Moles of $NH_3$ = $\frac{62.8kg/mole}{17g/mole}$ (since $Moles=\frac{mass of substance}{mass of one mole}$ )

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Moles of $CO_2$ = $\frac{105kg}{44g/mole}$

= $\frac{105000g}{44g/mole}$

= 2386.36 moles

Theoritically , moles of $NH_3$ required = double the moles of $CO_2$

but , $4011.76$ , the limiting reagent is $NH_3$

Theoritical moles of urea obtained = $\frac{1 mole CH_4N_2O}{2mole NH_3}\times4011.76 mole NH_3$

= $2005.88mole CH_4N_2O$

Mass of 2005.88 mole of $CH_4N_2O$ = $2005.88 mole \times\frac{60g CH_4N_2O}{1mole CH_4N_2O}$

= 120352.8g

$120352.8g\times \frac{1kg}{1000g}$

= 120.35kg

Therefore , theroritical yeild of urea = 120.35kg

Now , Percent yeild = $\frac{87.5kg}{120.35kg}\times100$

72.70%

Thus , the percent yeild for the reaction is 72.70%

8 0

3 years ago