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2 years ago

I have 2 Valence electrons and 1 energy levels

Chemistry

2 answers:

ladessa [460]2 years ago

8 0

Answer:

Hydrogen (H) and helium (He) have a valence shell containing one and two electrons respectively. They make up the first period (row) of the periodic table. Their valence electron/s are in the first energy level (n=1) , as is denoted by 1s1 and 1s2.

Explanation:

Logic... if its wrong imma cry

natima [27]2 years ago

3 0

Answer:

Hydrogen(H) and Heluim(He)

Explanation:

These are the only two valennce electrons and 1 energy levels.

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How to write a compound that contains one aluminum atom for every three chlorine atoms?

devlian [24]

AlCl₃
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4 0

3 years ago

The process of fermentation involves chemical reactions that convert solid glucose (C6H12O6) into aqueous ethanol and carbon dio

e-lub [12.9K]

Answer:

C6H12O6 —> 2C2H5OH + 2CO2

Explanation:

The equation for the reaction is given below:

C6H12O6 —> C2H5OH + CO2

We can balance the equation above as follow:

There are 12 atoms of H on the left side and 6 atoms of the right side. It can be balance by putting 2 in front of C2H5OH as shown below:

C6H12O6 —> 2C2H5OH + CO2

There are 6 atoms of C on the left side and 5 atoms on the right side. It can be balance by putting 2 in front of CO2 as shown below:

C6H12O6 —> 2C2H5OH + 2CO2

Now the equation is balanced.

4 0

2 years ago

Which solution below has the highest concentration of hydroxide ions?a. pH= 3.21b. pH= 7.00c. pH= 7.93d. pH= 12.59e. pH= 9.82

Sloan [31]

Answer:

<em>The solution that has the highest concentration of hydroxide ions is </em><u>d. pH = 12.59.</u>

Explanation:

You can solve this question using just some chemical facts:

pH is a measure of acidity or alkalinity: the higher the pH the lower the acidity and the higher the alkalinity.
The higher the concentration of hydroxide ions the lower the acidity or the higher the alkalinity of the solution, this is the higher the pH.

Hence, since you are asked to state the solution with the highest concentration of hydroxide ions, you just pick the highest pH. This is the option d, pH = 12.59.

These mathematical relations are used to find the exact concentrations of hydroxide ions:

pH + pOH = 14 ⇒ pOH = 14 - pH

pOH = - log [OH⁻] ⇒ $[OH^-]=10^{-pOH}$

Then, you can follow these calculations:

Solution pH pOH [OH⁻]

a. 3.21 14 - 3.21 = 10.79 antilogarithm of 10.79 = 1.6 × 10⁻¹¹

b. 7.00 14 - 7.00 = 7.00 antilogarithm of 7.00 = 10⁻⁷

c. 7.93 14 - 7.93 = 6.07 antilogarithm of 6.07 = 8.5 × 10⁻⁷

d. 12.59 14 - 12.59 = 1.41 antilogarithm of 1.41 = 0.039

e. 9.82 14 - 9.82 = 4.18 antilogarithm of 4.18 = 6.6 × 10⁻⁵

From which you see that the highest concentration of hydroxide ions is for pH = 12.59.

5 0

2 years ago

Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →

Inessa [10]

Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = $\frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }$

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

3 0

2 years ago

Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----> 2Cr3+ + 3Cl2 + 7H2O The reac

stich3 [128]

Answer:

Value of $Q_{p}$ for the given redox reaction is $1.0\times 10^{-8}$

Explanation:

Redox reaction with states of species:

$14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)$

Reaction quotient for this redox reaction:

$Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}$

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of $H_{2}O$ is taken as 1 due to the fact that $H_{2}O$ is a pure liquid.

$pH=-log[H^{+}]$

So, $[H^{+}]=10^{-pH}$

Plug in all the given values in the equation of $Q_{p}$ :

$Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}$

7 0

2 years ago