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Blababa [14]
3 years ago
14

a. silicon b. carbon c. beryllium d.chromium

Chemistry
2 answers:
Montano1993 [528]3 years ago
7 0

Answer:

It's a. silicon

Explanation:

zepelin [54]3 years ago
3 0
The answer is silicon because it’s atomic number is 14
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Consider the following reaction where Kc = 1.80×10-2 at 698 K:
Klio2033 [76]

Answer:

The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.

Explanation:

The reaction quotient Qc is a measure of the relative amount of products and reagents present in a reaction at any given time, which is calculated in a reaction that may not yet have reached equilibrium.

For the reversible reaction aA + bB⇔ cC + dD, where a, b, c and d are the stoichiometric coefficients of the balanced equation, Qc is calculated by:

Qc=\frac{[C]^{c}*[D]^{d}  } {[A]^{a}*[B]^{b}}

In this case:

Qc=\frac{[H_{2} ]*[I_{2} ] } {[HI]^{2}}

Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you have:

  • [H_{2} ]=\frac{2.09*10^{-2} moles}{1 Liter}=2.09*10⁻² \frac{moles}{liter}
  • [I_{2} ]=\frac{4.14*10^{-2} moles}{1 Liter}=4.14*10⁻² \frac{moles}{liter}
  • [I_{2} ]=\frac{0.280 moles}{1 Liter}= 0.280 \frac{moles}{liter}

So,

Qc=\frac{2.09*10^{-2} *4.14*10^{-2}  } {0.280^{2} }

Qc= 0.011

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

Being Qc=0.011 and Kc=1.80⁻²=0.018, then Qc<Kc. <u><em>The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.</em></u>

8 0
3 years ago
Does the motion of the moon affects how we see it from Earth? *
irina [24]
Yes that’s why we see it in different shapes all the time
7 0
2 years ago
Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
Manual says that he has reduce the amount of thermal energy in a cup of hot chocolate without lowering the temperature.How is th
ANEK [815]
Friction. It transforms other forms of energy into thermal energy
7 0
3 years ago
Four moles of Argon gas are at a pressure of 1.90 atm, and the temperature is 50 degrees C. What is the volume of the gas?
soldier1979 [14.2K]
N = 4 moles of Ar2, P = 1.90 atm, V = ?
T = 50C = 273 + 50K = 323K
PV = nRT --> V = nRT/P
V = (4)(.0821)(323)/1.90
V = 106.07/ 1.9
V = 55.8 L
4 0
2 years ago
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