(C) The silver carp is another species native to Asia that was introduced to aquaculture operations and sewage treatment ponds in the Mississippi River valley to control excess algae
accumulation. The species escaped from these operations into the Mississippi River and its tributaries and has the potential to invade the Great Lakes. Scientists are concerned that this
species is having severe negative impacts on these areas.
() Propose a potential solution, other than physically removing the carp from the rivers, to manage the problem of the invasive silver carp in the United States.

Answer:

Hydrogen H₂ will be the limiting reagent.

The excess reactant that will be left after the reaction is 3.45 moles.

4.3 moles of water can be produced.

Explanation:

The balanced reation is:

2 H₂ + O₂ → 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 1 mole of O₂ reacts with 2 moles of H₂, how much moles of H₂ will be needed if 5.6 moles of O₂ react?

[tex]moles of H_{2} =\frac{5.6 moles of O_{2} *2 mole of H_{2} }{1 mole of O_{2}}[/tex]

moles of H₂= 11.2 moles

But 11.2 moles of H₂ are not available, 4.3 moles are available. Since you have less moles than you need to react with 5.6 moles of O₂, hydrogen H₂ will be the limiting reagent and oxygen O₂ will be the excess reagent.

Then you can apply the following rules of three:

[tex]moles of O_{2} =\frac{1 mole of O_{2} *4.3 mole of H_{2} }{2 mole of O_{2}}[/tex]

moles of O₂= 2.15 moles

The excess reactant that will be left after the reaction can be calculated as:

5.6 moles - 2.15 moles= 3.45 moles

The excess reactant that will be left after the reaction is 3.45 moles.

[tex]moles of H_{2}O =\frac{2 moles of H_{2}O *4.3 mole of H_{2} }{2 mole of H_{2}}[/tex]

moles of H₂O= 4.3 moles

4.3 moles of water can be produced.

Answer:

A. 5.3 X 10^-2

Explanation:

I got 100 on PLATO

Explanation:

amount of baking soda = 2.01 g

---> 2.01 g NaHCO3 × (1 mol NaHCO3/84.007 g NaHCO3)

= 0.024 mol NaHCO3

mass of the product = 1.28 g

Possible reactions:

NaHCO3(s) ---> NaOH(s) + CO2(g)

0.024 mol NaHCO3 × (1 mol NaOH/1 mol NaHCO3)

= 0.024 mol NaOH × (39.997 g NaOH/1.mol NaOH)

= 0.96 g NaOH

2NaHCO3(s) ---> Na2O(s) + 2CO2(g) + H2O(g)

0.024 mol NaHCO3 × (1 mol Na2O/2 mol NaHCO3)

= 0.012 mol Na2O × (61.9789 g Na2O)

= 0.73 g Na2O

2NaHCO3(s) ---> Na2CO3(s) + CO2(g) + H2O(g)

0.024 mol NaHCO3 × (1 mol Na2CO3/2 mol NaHCO3)

= 0.012 mol Na2CO3 × (105.9888 g Na2CO3/1 mol Na2CO3)

= 1.27 g Na2CO3

Conclusion:

Based on the mass of the product in the crucible and comparing it with the predicted product mass in different reactions, I conclude that Na2CO3 is the most likely product in the crucible.

Equation balance:

2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

2Al(OH)3+3H2SO4

H=12

SO4=3

O=6

Al=2

Al2(SO4)3+6H2O

H=12

SO4=3

O=6

Al=2

Answer:

"When two or more elements combine to form more than one compound, the mass of one of them, which joins the fixed mass of the other, is in relation to whole numbers and simple" Example: The carbo

Answer:

4.17M

Explanation:

m1v1/m2v2=n1/n2

Answer:

friction

Explanation:

conduction doesn't really need much to start it. it basically has that reaction from the start so it's not conduction. Since heat is produced by friction in this situation it causes static to happen. so i think its friction

Answer:

the result is C and D. those are the ansers

Answer:C & D

Explanation:

Just took the test

Answer:

D . It describes a compound being broken down into the elements from which it was formed.

Explanation:

A compound consists of two or more elements chemically combined together in a chemical reaction. The compound carbon monoxide is represented by the chemical symbol, CO. This means that it contains 1 atom of the carbon element and 1 atom of the Oxygen element.

This compound is formed when carbon is directly oxidized in a limited supply of oxygen. In the above reaction, this compound is disintegrated into its individual components, namely, carbon and oxygen.

Answer:

1

Explanation:

Answer : A. 3

Explanation:

3Al(s) + 3O2(g) → 3Al2O3(s)

This is the right Answer :3

I hope you are having a great day :3

Explanation:

asexual reproduction involves 1 parent and produces offspring that are genetically identical to each other and parent.

Sexual reproduction involves 2 parents and produces offspring that are genetically unique.

Answer:

II would say there is co2 so i would say 30.4

Explanation:

Hope this helps

Answer:

250cm³ de disolución se deben preparar

Explanation:

El porcentaje volumen volumen (%v/v) se define como el volumen de soluto -En este caso, H2O2- en 100cm³ de disolución.

Se desea preparar una solución que contenga 20cm³ de H2O2 por cada 100cm³

Como solamente se cuenta con 50cm³ de H2O2, el volumen que se debe preparar es:

50cm³ H2O2 * (100cm³ Disolución / 20cm³ H2O2) =

Answer:

To find the relative formula mass (M r) of a compound, you add together the relative atomic mass values (A r values) for all the atoms in its formula.

Explanation: Find the M r of carbon monoxide, CO. Find the M r of sodium oxide, Na 2O. The relative formula mass of a substance, shown in grams, is called one mole of that substance.

Answer:

100 L

Explanation:

Use Boyle's Law, p1v1 = p2v2.

p1 = 4 atm

v1 = 50L

p2 = 2 atm

v2 = ?

p1v1 = p2v2

(4 atm) (50L) = (2 atm) (v2)

v2 = (4 atm)(50L) / (2 atm) = 100 L

Answer:

Option C (1.2 L) is the correct option.

Explanation:

The given values are:

Pressure,

[tex]P_1=1.9 \ atm[/tex]

[tex]P_2=3.8 \ atm[/tex]

Volume,

[tex]V_1=2.3 \ L[/tex]

[tex]V_2=?[/tex]

As we know,

The gas equation is:

⇒  [tex]PV=nRT[/tex]

At constant temperature,

⇒  [tex]P_1 V_1=P_2 V_2[/tex]

or,

⇒      [tex]V_2=\frac{P_1 V_1}{P_2}[/tex]

By substituting the values, we get

⇒           [tex]=\frac{1.9\times 2.3}{3.8}[/tex]

⇒           [tex]=\frac{4.37}{3.8}[/tex]

⇒           [tex]=1.15 \ L[/tex]

or,

⇒           [tex]=1.2 \ L[/tex]

Answer:

Electrons and protons are not positively charged

Explanation:

Neutrons are the only ones positively charged in the structure of an atom.... Also like charges repel which means a -ive and a - ive will repel when together

Answer:

The arrangement of electrons within an atom is called electron configuration

If I help you, will you help other people?

Actually "this is the worst question i've ever seen use the internet bud" why don't you do that.

Answer:

D

Explanation:

Because it has equal number of elements on each side of the arrow

Answer:

a) pentane

b) 1,3-dimethyl butane

c) 3,4-diethyl-5-methlyl heptane

d) 1,3,5-triethyl-2,4,6-trimethyl  cyclohexane

Explanation:

The essence of IUPAC nomenclature is to provide a generally accepted nomenclature for chemical compounds which enables the structure of a compound to be drawn from its name without ambiguity.

In order to name straight chain organic compounds, we have to identify the parent chain and then give numbers to the substituents or locants such that they have the lowest possible numbers in alphabetical order.

The correct names of the organic compounds in the question are;

a) pentane

b) 1,3-dimethyl butane

c) 3,4-diethyl-5-methlyl heptane

d) 1,3,5-triethyl-2,4,6-trimethyl cyclohexane

Answer:

9 protons

Explanation:

By looking at the periodic table, you will see that Fluorine has 9 protons. Since the number of electrons equal the number of protons, Fluorine has 9 electrons as well. Meanwhile, it's mass number of 19, minus 10 neutrons, gives you 9 protons or electrons. Hence, the atom would be Fluorine.

hope u make me brainlesst ʘ‿ʘ

Answer:

i think so for this i will 9 protons . atomic mass is 19. 10 neutrons

Answer:

water vapor (gas) is becoming liquid

Answer:

2Cu2O + C --> 4Cu + CO2

Explanation

The Law of conservation of Mass states You can not destroy or create math in a chemical reaction, meaning you must have the same amount at the beginning and the end. This equation keeps the same amount of atoms on both sides.

Answer: Yes, it is a balanced equation.

Explanation:

https://chemequations.com/en/?s=Cu2O+%2B+C+%3D+Cu+%2B+CO2

This is where I got the answer so you know.  Hope this helps :)

It's ammonia (NH3).....

Answer:

A molecule with one lone pair and three bond pair

- Ammonia

Answer:

[tex]tiempo=0.72s[/tex]

Explanation:

¡Hola!

En este caso, como podemos observar, la relación entre la rapidez de reaction, las moles y el tiempo de reacción es:

[tex]rapidez=\frac{mol}{tiempo}[/tex]

De este modo, despejamos el tiempo en esta ecuación para obtener:

[tex]tiempo=\frac{mol}{rapidez}[/tex]

Así, reemplazamos la rapidez y moles dadas para obtener:

[tex]tiempo=\frac{0.323mol}{0.450mol/s}\\\\tiempo=0.72s[/tex]

¡Saludos!

Answer:

0.702 /s

Explanation:

Rate constant at [tex][298 \mathrm{~K}, \mathrm{~K}_{1}=3.46 \times 10^{-2} \mathrm{~s}^{-1}[/tex]

Rate constant at [tex]350 \mathrm{~K}, \mathrm{~K}_{2}=?[/tex]

[tex]T_{1}=298 \mathrm{~K}[/tex]

[tex]T_{2}=350 \mathrm{~K}[/tex]

Activation energy, [tex]\mathrm{Ea}=50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}[/tex]

Use the following equation to calculate [tex]K_{2}$ at $350 \mathrm{~K}[/tex]

Use the following equation to calculate [tex]K_{2}$ at $350 \mathrm{~K}[/tex]

[tex]\ln \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right][/tex]

Therefore,

 [tex]\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{1}{298 \mathrm{~K}}-\frac{1}{350 \mathrm{~K}}\right][/tex]

[tex]\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{52 \mathrm{~K}}{298 \mathrm{~K} \times 350 \mathrm{~K}}\right][/tex]

[tex]\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}} &=\mathrm{e}^{3.01}[/tex]

[tex]\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}} &=20.3[/tex]

[tex]K_{2} &=20.3 \times 3.46 \times 10^{-2} \mathrm{~s}^{-1}[/tex]

[tex]&=0.702 \mathrm{~s}^{-1}[/tex]

hence, the rate constant at [tex]350 \mathrm{~K} [/tex] is 0.702[tex]\mathrm{~s}^{-1}[/tex]