All categories

3 years ago

When the football is on its way up, 1. BLANK energy is being transformed into 2. BLANK energy.

Chemistry

2 answers:

weqwewe [10]3 years ago

8 0

When the football is on its way up, Kinetic energy is being transformed into Gravitational Potential energy.

oksian1 [2.3K]3 years ago

6 0

I think it is 1 because as it goes up it loses energy and when it comes down it is gaining that energy back.

You might be interested in

A solution of K2SO4 and KCl is added to a solution of Ba(NO3)2. Which of these compounds will precipitate out of this combined s

Archy [21]

A solution of K2SO4 and KCl is added to a solution of Ba(NO3)2. BaS0₄ (s) will precipitate out of this combined solution.

Molecular equation

K2SO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + 2 KNO3(aq)?

This equation represents a double displacement (replacement) reaction, also called a metathesis reaction, in which the reactant ions exchange places to form new products. The general equation is:

A-B + C-D → A-D + C-B;

where A and C are cations, and B and D are anions.

Complete ionic equation: Includes all ions and the precipitate.

2K^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2[NO3]^-(aq) → 2K^+(aq) + 2[NO3]^- + BaSO4(s)

In an aqueous solution, precipitation is the process of transforming a dissolved substance into an insoluble solid from a super-saturated solution.

The solid formed is called the precipitate. In case of an inorganic chemical reaction leading to precipitation, the chemical reagent causing the solid to form is called the precipitant.

Learn more about precipitation here : brainly.com/question/1783904

#SPJ4

6 0

1 year ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago

I need help on this one pliz

Artyom0805 [142]

The answer would be 16.04.

8 0

3 years ago

Which of the following reactions could be an elementary reaction? 2 NO2(g) + F2(g) → 2NO2F(g) Rate = k[NO2][F2] H2(g) + Br2(g) →

podryga [215]

Answer: The correct answer is $NO(g)+O_2(g)\rightarrow NO_2(g)+O(g);Rate=k[NO][O_2]$

Explanation:

Molecularity of the reaction is defined as the number of atoms, ions or molecules that must colloid with one another simultaneously so as to result into a chemical reaction.

Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.

Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.

For the given reactions:

Equation 1: $2NO_2(g)+F_2(g)\rightarrow 2NO_2F(g);Rate=k[NO_2][F_2]$

Molecularity of the reaction = 2 + 1 = 3

Order of the reaction = 1 + 1 = 2

This is not considered as an elementary reaction.

Equation 2: $H_2(g)+Br_2(g)\rightarrow 2HBr(g);Rate=k[H_2][Br_2]^{1/2}$

Molecularity of the reaction = 1 + 1 = 2

Order of the reaction = $1+\frac{1}{2}=\frac{3}{2}$

This is not considered as an elementary reaction.

Equation 3: $NO(g)+O_2(g)\rightarrow NO_2(g)+O(g);Rate=k[NO][O_2]$

Molecularity of the reaction = 1 + 1 = 2

Order of the reaction = 1 + 1 = 2

This is considered as an elementary reaction.

Equation 4: $NO_2(g)+CO(g)\rightarrow NO(g)+CO_2(g);Rate=k[NO_2]^2$

Molecularity of the reaction = 1 + 1 = 2

Order of the reaction = 2 + 0 = 2

In this equation, the order with respect to each reactant is not equal to its stoichiometric coefficient which is represented in the balanced chemical reaction. Hence, this is not considered as an elementary reaction.

Hence, the correct answer is $NO(g)+O_2(g)\rightarrow NO_2(g)+O(g);Rate=k[NO][O_2]$

3 0

3 years ago

using the equation, c5h12 8o2 arrow 5co2 6h2o, if 108 g of water are produced, how many grams of oxygen were consumed?

vfiekz [6]

Molar mass:

H₂O = 18.0 g/mol

O₂ = 32.0 g/mol

C₅H₁₂ + 8 O₂ -> 5 CO₂ + 6 H₂O

8 x (32 g ) ------------ 6 x (18 g )
mass O₂ ------------ 108 g H₂O

mass O₂ = 108 x 8 x 32 / 6 x 18

mass O₂ = 27648 / 108

mass O₂ = 256 g

hope this helps!

8 0

2 years ago