All categories

3 years ago

40 POINTS write a lab report about types of reactions

Chemistry

2 answers:

Masteriza [31]3 years ago

6 0

Answer:

The purpose of this lab was to explore the different types of chemical reactions in this laboratory procedure. The question for this lab is, how does knowing the reactants and products help me classify a chemical reaction. In this experiment, I made observations of the reactants and the products. I used these observations to write the balanced chemical equations and to classify the reaction it made.

Explanation:

frozen [14]3 years ago

5 0

Answer:

bruh

Explanation:

You might be interested in

What is the cell potential of an electrochemical cell that has the half-reactions shown below?

Karo-lina-s [1.5K]

Answer:

E°(Ag⁺/Fe°) = 0.836 volt

Explanation:

3Ag⁺ + 3e⁻ => Ag°; E° = +0.800 volt

Fe° => Fe⁺³ + 3e⁻ ; E° = -0.036 volt

_________________________________

Fe°(s) + 3Ag⁺(aq) => Fe⁺³(aq) + 3Ag°(s) ...

E°(Ag⁺/Fe°) = E°(Ag⁺) - E°(Fe°) = 0.800v - ( -0.036v) = 0.836 volt

4 0

3 years ago

Assume that 1.0 mol of C4H10 is completely burned in excess oxygen to form carbon dioxide and water. How many moles of CO2 would

In-s [12.5K]

You need to first write a chemical equation and balance it
C₄H₁₀ + O₂ → CO₂ + H₂O
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
1.0 moles X moles
1.0 mol C₄H₁₀ ( $\frac{8 mol CO₂}{2 mol C₄H₁₀}$ ) = 4 moles of CO₂

7 0

3 years ago

Why does it generally take more enthalpy to ignite a solid than a gas or liquid?

Anuta_ua [19.1K]

Answer:

It is due to the nature of the reactants

Explanation:

To ignite a solid, we require more heat component compared to liquids and gases. For ignition to occur, oxygen gas combines with a reactant in most cases.

Some factors affect the rate rate at which a chemical proceeds. One of the factors is the nature of reactants.

The solid phase is very slow while the gaseous phase is rapid and fast.

solid phase < liquid phase < gas phase

Gases are free and the molecules move in all direction. They easily combine and react very fast.

6 0

3 years 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

The Michaelis–Menten equation is an expression of the relationship between the initial velocity ????0 of an enzymatic reaction a

hammer [34]

1 doubling s will almost double the rate

8 0

3 years ago