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Jlenok [28]
3 years ago
6

The equilibrium constant for the reaction: HCHO(g) ⇌ H2(g) + CO(g).

Chemistry
1 answer:
tiny-mole [99]3 years ago
3 0

Answer:

K = [H2] [CO] / [HCHO]

Explanation:

HCHO(g) ⇌ H2(g) + CO(g)

We can obtain the expression for the equilibrium constant for the above equation as follow:

Equilibrium constant, K for a given reaction is the ratio of the concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

Thus, the equilibrium constant, K for the above equation can be written as follow:

K = [H2] [CO] / [HCHO]

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Which of the following combination of elements would result in covalent compound? * W X Y Z Vand X Wand Z Y and Z Wand y​
TEA [102]

Answer:

C. Y & Z

Explanation:

V, W are imaginary metals here because their valence electrons are typically less than 4. X, Y, Z are non-metals and have higher valence electrons. Here, if V or W bind with X, Y, or Z we make ionic bond (because metal + non metal = ionic). But, if X binds with Y or Z or any combinations of any two of the three non-metals results in covalent bond (non metal + non metal = covalent).

Thus, Y and Z make covalent.

4 0
2 years ago
Calculate the number of moles in 7.41 X 1017 atoms Ag.
jolli1 [7]

Answer:

1.23x10^-6 mole

Explanation:

A clear understanding of Avogadro's hypothesis proved that 1mole of any substance contains 6.02x10^23 atoms. This indicates that 1mole of Ag contains 6.02x10^23 atoms.

Now, 1f 1mole of Ag contains 6.02x10^23 atoms, then Xmol of Ag will contain 7.41x10^17 atoms i.e

Xmol of Ag = 7.41x10^17/6.02x10^23 = 1.23x10^-6 mole

4 0
3 years ago
A buffer solution is composed of 4.00 4.00 mol of acid and 3.25 3.25 mol of the conjugate base. If the p K a pKa of the acid is
Reika [66]

<u>Answer:</u> The pH of the buffer is 4.61

<u>Explanation:</u>

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})

We are given:

pK_a = negative logarithm of acid dissociation constant of weak acid = 4.70

[\text{conjuagate base}]} = moles of conjugate base = 3.25 moles

[\text{acid}]  = Moles of acid = 4.00 moles

pH = ?

Putting values in above equation, we get:

pH=4.70+\log(\frac{3.25}{4.00})\\\\pH=4.61

Hence, the pH of the buffer is 4.61

8 0
3 years ago
Complete combustion of 5.90 g of a hydrocarbon produced 18.8 g of CO2 and 6.75 g of H2O.
Inessa05 [86]
First we have to find moles of C:
Molar mass of CO2:
12*1+16*2 = 44g/mol
(18.8 g CO2) / (44.00964 g CO2/mol) x (1 mol C/ 1 mol CO2) =0.427 mol C 
Molar mass of H2O:
2*1+16 = 18g/mol
As there is 2 moles of H in H2O,
So,

<span>(6.75 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) = 0.74mol H </span>

<span>Divide both number of moles by the smaller number of moles: </span>
<span>As Smaaler no moles is 0.427:
So,
Dividing both number os moles by 0.427 :
(0.427 mol C) / 0.427 = 1.000 </span>
<span>(0.74 mol H) / 0.427 = 1.733 </span>

<span>To achieve integer coefficients, multiply by 2, then round to the nearest whole numbers to find the empirical formula: 
C = 1 * 2 = 2
H = 1.733 * 2 =3.466
So , the empirical formula is C2H3</span>
4 0
3 years ago
Read 2 more answers
The decomposition of carbon disulfide to carbon monosulfide and sulfur is first order with k=2.8 ×10^-7 at 1000°C .What is the h
RoseWind [281]

Answer:

2.5×10⁶ s

Explanation:

From the question given above, the following data were obtained:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

The half-life of a first order reaction is given by:

Half-life (t½) = 0.693 / Rate constant (K)

t½ = 0.693 / K

With the above formula, we can obtain the half-life of the reaction as follow:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

t½ = 0.693 / K

t½ = 0.693 / 2.8×10¯⁷

t½ = 2.5×10⁶ s

Therefore, the half-life of the reaction is 2.5×10⁶ s

6 0
2 years ago
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