Use scientific notation to express the number 355000 to:

Help ASAP!!! I need it now i have to write similar test tmrw

dusya [7]

A is Ea, which stands for activating energy. Energy is needed to get the reaction underway and Ea is the energy needed to “start” the reaction.

B is the temperature either released or absorbed.

The diagram shows that the reaction is exothermic based on the fact that the products energy is lower than the reactants. That is because energy (which is temperature in this case) is released during the process. If the reactants would have been lower than the products, the reaction would be endothermic.

3 0

2 years ago

Read 2 more answers

HELP !!!!!!! Please !!!!!!!

DerKrebs [107]

Answer:

I guess 4 weeks is a little late.

Explanation:

7 0

3 years ago

At a certain concentration of H2 and NH3, the initial rate of reaction is 0.120 M / s. What would the initial rate of the reacti

mel-nik [20]

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

$rate=k[H_2]^2[NH_3]$

At a certain concentration of $H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]$

As we are given that:

Initial rate = 0.120 M/s

Expression for rate law for first observation:

$0.120=k[H_2]^2[NH_3]$ ....(1)

Expression for rate law for second observation:

$R=k(\frac{[H_2]}{2})^2[NH_3]$ ....(2)

Dividing 2 by 1, we get:

$\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}$

$\frac{R}{0.120}=\frac{1}{4}$

$R=0.03M/s$

Therefore, the initial rate of the reaction will be, 0.03 M/s

5 0

3 years ago

Which of the following oils would consume the greatest number of equivalents of hydrogen when subject to catalytic hydrogenation

Murljashka [212]

Answer: Option (B) is the correct answer.

Explanation:

When a fatty acid contains high number of double bonds then its unsaturation will also be high and hence, it will consume greater number of equivalents of hydrogen.

In corn oil, there are no unsaturated sites are present.

In olive oil, there is one unsaturated site with majority of oleic acid. In olive oil, there are more than 70% of total unsaturated oils.

In lard oil, there are around 60% of unsaturated oils.

In herring oil, there are highest number of saturated fatty acids and lowest polyunsaturated acids.

Thus, we can conclude that out of the given options, olive oils would consume the greatest number of equivalents of hydrogen when subject to catalytic hydrogenation.

5 0

3 years ago

A sample of sodium sulfite has a mass of 2.80 g.

Gnom [1K]

Answer:

For a: The number of sodium ions in given amount of sodium sulfite are $2.65\times 10^{22}$

For b: The number of sulfite ions in given amount of sodium sulfite are $1.325\times 10^{22}$

For c: The mass of one formula unit of sodium ions is $2.09\times 10^{-22}$ grams

Explanation:

The chemical formula of sodium sulfite is $Na_2SO_3$ . It is formed by the combination of 2 sodium $(Na^+)$ ions and 1 sulfite $(SO_3^{2-})$ ions

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of sodium sulfite = 2.80 g

Molar mass of sodium sulfite = 126 g/mol

Putting values in above equation, we get:

$\text{Moles of sodium sulfite}=\frac{2.80g}{126g/mol}=0.022mol$

For a:

Moles of sodium ions in sodium sulfite = (2 × 0.022) moles

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, 0.022 moles of sodium sulfite will contain = $(2\times 0.022\times 6.022\times 10^{23})=2.65\times 10^{22}$ number of sodium ions

Hence, the number of sodium ions in given amount of sodium sulfite are $2.65\times 10^{22}$

For b:

Moles of sulfite ions in sodium sulfite = (1 × 0.022) moles

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, 0.022 moles of sodium sulfite will contain = $(1\times 0.022\times 6.022\times 10^{23})=1.325\times 10^{22}$ number of sulfite ions

Hence, the number of sulfite ions in given amount of sodium sulfite are $1.325\times 10^{22}$

For c:

Molar mass of sodium sulfite = 126 g/mol

According to mole concept:

$6.022\times 10^{23}$ number of formula units are present in 1 mole of a compound

Or, $6.022\times 10^{23}$ number of formula units of sodium sulfite have a mass of 126 grams

So, 1 formula unit of sodium sulfite will have a mass of = $\frac{126}{6.022\times 10^{23}}\times 1=2.09\times 10^{-22}g$

Hence, the mass of one formula unit of sodium ions is $2.09\times 10^{-22}$ grams

7 0

3 years ago