A doubling of the concentration of a doubles the rate of the reaction. true or false

Which is the most important question for deciding if a chemical reaction has occurred?

777dan777 [17]

Answer:

all these are physical properties except release of heat so it's probably heat energy given off

7 0

3 years ago

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2C2H6 + 7O2 ------> 4CO2 + 6H2O

Romashka-Z-Leto [24]

We convert the masses of our reactants to moles and use the stoichiometric coefficients to determine which one of our reactants will be limiting.

Dividing the mass of each reactant by its molar mass:

(10 g C2H6)(30.069 g/mol) = 0.3326 mol C2H6

(10 g O2)(31.999 g/mol) = 0.3125 mol O2.

Every 2 moles of C2H6 react with 7 moles of O2. So the number of moles of O2 needed to react completely with 0.3326 mol C2H6 would be (0.3326)(7/2) = 1.164 mol O2. That is far more than the number of moles of O2 that we are given: 0.3125 moles. Thus, O2 is our limiting reactant.

Since O2 is the limiting reactant, its quantity will determine how much of each product is formed. We are asked to find the number of grams (the mass) of H2O produced. The molar ratio between H2O and O2 per the balanced equation is 6:7. That is, for every 6 moles of H2O that is produced, 7 moles of O2 is used up (intuitively, then, the number of moles of H2O produced should be less than the number of moles of O2 consumed).

So, the number of moles of H2O produced would be (0.3125 mol O2)(6 mol H2O/7 mol O2) = 0.2679 mol H2O. We multiply by the molar mass of H2O to convert moles to mass: (0.2679 mol H2O)(18.0153 g/mol) = 4.826 g H2O.

Given 10 grams of C2H6 and 10 grams of O2, 4.826 g of H2O are produced.

8 0

2 years ago

A sample of water is heated from room temperature to just below the boiling point. The overall change in temperature is 72Â°C. E

Jlenok [28]

Answer : The correct option is, (a) 345 K

Explanation :

The conversion used for the temperature from degree Celsius to Kelvin is:

$K=273.15+^oC$

where,

$K$ = temperature in Kelvin

$^oC$ = temperature in centigrade

As we are given the temperature in degree Celsius is, 72

Now we have to determine the temperature in Kelvin.

$K=273.15+^oC$

$K=273.15+(72^oC)$

$K=345.15\approx 345$

Therefore, the temperature in Kelvin is, 345 K

7 0

3 years ago

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At 25°C, the equilibrium constant Kc for the reaction 2A(aq) ↔ B(aq) + C(aq) is 65. If 2.50 mol of A is added to enough water to

Svet_ta [14]

Answer:

0.146 M

Explanation:

Equation for the reaction :

2A(aq) ↔ B(aq) + C(aq)

$K_c$ = 65

Molar concentration of A = $\frac{2.50 mol}{1.00 L}$

= 2.5 M

2A(aq) ↔ B(aq) + C(aq)

Initial 2.50 0 0

Change - 2x + x + x

Equilibrium 2.50 - 2x +x +x

$K_c =\frac {[B][C]}{[A]^2}$

$65 = \frac{[x][x]}{[2.5-2x]^2}$

$65 = \frac{[x]^2}{[2.5-2x]^2}$

$65 = (\frac{[x]}{[2.5-2x]})^2$

$\sqrt 65 = \sqrt {(\frac{[x]}{[2.5-2x]})^2}$

$8.062 = \frac{x}{2.5-2x}$

8.062(2.5 - 2x) = x

20.155 - 16.124x = x

20.155 = 16.124x+x

20.155 = 17.124x

x = $\frac{20.155}{17.124}$

x = 1.177

[A] = 2.5 - 2x

= 2.5 - 2(1.177)

= 0.146 M

Therefore, the equilibrium concentration of A = 0.146 M

4 0

3 years ago

A mixture of carbon dioxide and hydrogen gases is maintained in a 6.68 L flask at a pressure of 2.14 atm and a temperature of 19

matrenka [14]

Answer:

The mass of hydrogen gas in the mixture: w₂ = 0.433 g

Explanation:

According to the ideal gas equation:

for an ideal gas, $P.V = n_{total}.R.T$

and $n_{total}= n_{1}+n_{2}$

Here, P: total pressure of the gases = 2.14 atm

V: total volume of the gases = 6.68 L

T: temperature = 19 °C = 19+273.15 = 292.15K (∵ 0°C = 273.15K)

R: gas constant = 0.08206 L·atm·K⁻¹·mol⁻¹

$n_{total}$ : total number of moles of gases

To calculate the total number of moles of gases:

$n_{total} = \frac{P.V}{R.T} = \frac{2.14 atm\times 6.68 L}{0.08206 LatmK^{-}mol^{-}\times 292.15K}$ = 0.5963 moles

Let, the number of moles of carbon dioxide be n₁ and number of moles of hydrogen be n₂

Given: mass of carbon dioxide: w₁ = 16.8 g, mass of hydrogen: w₂ = ?g

molar mass of carbon dioxide: m₁ = 44.01 g/mol, molar mass of hydrogen: m₂= 2.016 g/mol

Therefore, $n_{total}$ = n_{1}+n_{2} = (w₁ ÷ m₁) + (w₂ ÷ m₂)

⇒ 0.5963 mol = (16.8 g ÷ 44.01 g/mol) + (w₂ ÷ 2.016 g/mol)

⇒ 0.5963 mol = (0.3817mol) + (w₂ ÷ 2.016 g/mol)

⇒ 0.5963 mol - 0.3817mol = (w₂ ÷ 2.016 g/mol)

⇒ 0.2146 mol = (w₂ ÷ 2.016 g/mol)

⇒ w₂ = 0.433 g

Therefore, the mass of hydrogen gas in the mixture: w₂ = 0.433 g

4 0

3 years ago