Plants use water to make sugar during the process of transpiration true or false please help out thanks

For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The

Alik [6]

Answer : The rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

Explanation :

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:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

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

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

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

Putting values in equation 1, we get:

$6.0\times 10^{-4}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-2}M^{-2}s^{-1}$

Thus, the value of the rate constant 'k' for this reaction is $7.5\times 10^{-2}M^{-2}s^{-1}$

Now we have to calculate the rate for trial 5 that starts with 0.90 M of reagent A, 0.60 M of reagents B and 0.70 M of reagent C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-2})\times (0.90)^2(0.60)^0(0.70)^1$

$\text{Rate}=4.25\times 10^{-2}Ms^{-1}$

Therefore, the rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

3 0

3 years ago

An atom of an element has 13 electrons what element is it

balandron [24]

Answer:

Aluminium

Explanation:

Aluminium is the only atom that has 13 electron

7 0

3 years ago

Suppose 0.410 kg of hexane are burned in air at a pressure of exactly 1 atm and a temperature of 13.0 °C. Calculate the volume o

3241004551 [841]

Answer:

The answer is 671 litres of carbon dioxide is produced from 0.410 kg of hexane

Explanation:

We first write a balanced reaction for the complete combustion of hexane thus

The stoichiometry of the cumbustion of hexane in air is

2C6H14(g)+18O2(g)→12CO2(g)+14H2O(l) or

C6H14(g)+9O2(g)→6CO2(g)+7H2O(l)

From the above reaction it is observed that one mole of hexane burns completely in the presence of oxygen to produce 6 moles of carbon dioxide

Therefore we calculate the nuber of moles of hexane present in the sample thus

Mass hexane of sample = 0.41 kg

Molar nass of hexane = 86.18 g/mol

number of moles of hexane = (mass of hexane)/(molar mass of hexane) = (0.41×1000)/86.16 = 410/86.16 = 4.76 moles

As we have seen from the chemical reaction, 1 mole of H6H14 produces 6 moles of CO2 hence 4.76 moles of Hexane produces

4.76×6 moles of CO2 which is 28.55 moles of CO2

From the question we have the temperature and the pressure of the production of CO2 as

Temperature of reaction = 13° C converting to kelving gives= 13+273.15 = 286.15 K

and pressure = 1 atmosphere or 101325 Pa

13.0∘C=13.0∘C+273.15=286.15 K

The volume of the produced CO2 can be calculated using the combined ideal gas equation given by

P×V=n×R×T where

Here

P = Gas pressure (of CO2 )

V = Volume (of the CO2)

n = number of moles of gas (CO2) present

R = universal gas constant, equal to 0.0821 atm× L/(mol× K )

T = absolute temperature in Kelvin

Thus we have

1×V = 28.55×0.0821×286.15 or V = 670.76L

Rounding up the answer to 3 significant digits we have

670.76L ≅ 671L

671 litres of carbon dioxide is produced from 0.410 kg of hexane

8 0

4 years ago

Under which conditions would the solubility of a gas be greatest? high pressure and high temperature high pressure and low tempe

kiruha [24]

Option B: high pressure and low temperature

A gas is more soluble under high pressure and low temperature conditions.

On increasing temperature of a gas, its kinetic energy increases. The increase in kinetic energy increases the motion of particles of gas this causes most of the gaseous particles to escape from the gas phase. Thus, less particles are available to dissolve in liquid and solubility decreases.

The effect of pressure on solubility of gas can be explained with the help of Henry's law. According to the law, at constant temperature, solubility of gas and partial pressure of gas are related to each other as follows:

$p=k_{h}c$

Here, p is the partial pressure of the gas, $k_{h}$ is Henry's law constant, and

c is the concentrate of the gas.

According to above relation, concentration of gas decreases on decreasing partial pressure. Thus, on increasing pressure, concentration of gas increases this increases the solubility of gas in liquid.

Therefore, solubility of gas is greatest at high pressure and low temperature.

4 0

4 years ago

Read 2 more answers

PLEASE HELP I WILL GIVE BRAINLYEST!!!!

adoni [48]

I think that it is true to the bone

8 0

3 years ago