What is the mass of 8.12 ×10^23

Answer:

18.45 g of C

Explanation:

This is a problem of rules of three:

1 mol of C₃H₈ contains 3 moles of C and 8 moles of H

If 8 moles of H are contained in 1 mol of propane

4.10 moles of H are contained in (4.1 . 1) /8 = 0.5125 moles

Now, If 1 mol of propane contains 3 moles of C

0.5125 moles of propane may contain (0.5125 . 3) / 1 = 1.5375 moles of C

Let's convert the moles to mass:

1.5375 mol . 12 g /mol = 18.45 g

Answer:

Explanation:

Hope this helps

Answer:

Light can be treated like particles because it is made of chunks like things called protons.

Answer:

The solute is salt and the solvent is water

Explain:

Because salt is a component the dissolves in the solvent

Answer:

b. +3 and +6.

Explanation:

Zeff = Z - S

The Z denotes the no of protons i.e. atomic number

S denotes the non-valence electrons

For boron,

the electronic configuration is 1s₂ 2s₂ 2p₄

Now  

Z = 5, S = 2

So,  

Zeff = 5-2

= +3

For O, the electronic configuration is 1s₂ 2s₂ 2p₄

So,  

Z = 8, S = 2

= 8-2

= +6

Hence, the second option is correct

Answer:

I assume your talking about carbon when you say 12 so it'd be 12 grams if you are

Explanation:

The molar mass of any substance in grams per mole is numerically equal to the mass of that substance expressed in atomic mass units.

Hope this helps you some

Equilibrium shifts to the right.

OPTION A

Where is ur question ⁉️

Answer: The concentration of the solution is 55.66g[tex]KNO_3[/tex]/100g[tex]H_2O[/tex]

Explanation:

A solution consists of solute and solvent. A solute is defined as the component present in a smaller proportion while the solvent is defined as the component that is present in a larger proportion.

We are given:

Mass of [tex]KNO_3[/tex] = 2.31 g

Mass of water = 4.15 g

To calculate the concentration in g[tex]KNO_3[/tex]/100 g [tex]H_2O[/tex], we apply unitary method:

In 4.15 g of water, the mass of [tex]KNO_3[/tex] present is 2.31 g

So, in 100 g of water, the mass of [tex]KNO_3[/tex] present will be [tex]\frac{2.31}{4.15}\times 100=55.66g[/tex]

Hence, the concentration of the solution is 55.66g[tex]KNO_3[/tex]/100g[tex]H_2O[/tex]

Answer: In order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.

Explanation:

More is the concentration of reactant molecules more will be the number of collisions between their molecules. As a result, more readily the products will be formed.

Hence, for the given reaction when concentration of HCl is increased then there will be increase in the number of collisions between reactants.

Thus, we can conclude that in order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.

Answer:

I don't no answer sorry

Explanation:

you follow me

Answer:

Chromosomes and I think its too many

Explanation:

Answer:

Wearing gloves

Explanation:

it helps on not touching chemicals

The question is incomplete, the complete question is;

1.000 atm of Oxygen gas, placed in a container having apinhole opening in its side. leaks from the container 2.14 timesfaster thatn 1.000 atm of an unknown gas placed in this sameapparatus. Which of the following species could be theunknown gas?

A. CL2

B. SF6

C. Kr

D. UF6

E. Xe

Answer:

SF6

Explanation:

From Graham's law;

Let the rate of diffusion of oxygen be R1

Let the rate of diffusion of unknown base be R2

Let the molar mass of oxygen by M1

Let the molar mass of unknown gas be M2

Hence;

R1/R2 = √M2/M1

So;

2.14/1 = √M2/32

(2.14/1)^2 = M/32

M= (2.14/1)^2 × 32

M= 146.6

This is the molar mass of SF6 hence the answer above.

Answer:

the correct answer is four decimal places

Answer:

2

Explanation:

You write your answer with the same number of significant figures as the number with the smallest amount of figures, in this case, that number is 2.1 so your answer should be written with 2 significant figures.

Answer:

all of the above

Explanation:

have a nice day :-)

Answer:

[tex]molar \: mass \: of \: copper(ii)nitrate = 64 + (14 + 48) \times 3 \\ = 250 \: g \\ 64 \: g \: of \: copper \: produces \: 250 \: g \: of \: copper \: nitrate \\ 10.36 \: g \: of \: copper \: will \: produce \: ( \frac{10.36 \times 250}{64} ) \: g \\ = 40.7 \: g[/tex]

Answer:

Explanation:

From the graphical diagram attached below; we can see the relationship between the concentration of [tex]H_2O_2[/tex] which declines exponentially in relation to the time and it obeys the equation: [tex]\mathtt{y = 0.9951 e^{-8\times 10^{-4}x}}[/tex]

This relates to the 1st order reaction rate, whereby:

The integrated rate law[tex]\mathtt{ [A] = [A]_o e^{-kt}}[/tex]

here:

[A] = reactant concentration at time (t)

[A]_o = initial concentration for the reactant

k = rate constant

As such, the order of the reaction is the first order

Rate constant [tex]\mathtt{k = 8\times 10^{-4} {s^{-1}}}[/tex]

Rate law [tex]\mathtt{= k[H_2O_2]}[/tex]

The integrated rate law [tex]\mathtt{[H_2O_2] = [H_2O_2]_oe^{-(8*10^{-4})t}}[/tex]

From the given table:

the initial concentration of [tex]H_2O_2[/tex] = 1.00 M

∴

We can determine the concentration of the reactant at 4000s by using the formula:

[tex]\mathtt{[H_2O_2] = [H_2O_2]_oe^{-8*10^{-4}(t)}}[/tex]

[tex]\mathtt{[H_2O_2] = (1.00\ M)*e^{-8*10^{-4}(4000)\ s}}[/tex]

[tex]\mathtt{[H_2O_2] =0.0407 \ M}[/tex]

Finally, at 4000s: the average rate is:

[tex]\mathtt{= (8*10^{-4} \ s^{-1})(4000 \ s) }\\ \\ \mathtt{ = 3.256 \times 10^{-5} \ M/s}[/tex]

Answer:

Volume of the reaction vessel is increased - shift to the left

The reaction is cooled down - shift to the right

H2 is added to the system - shift to the right

The pressure of the system is decreased - shift to the left

A catalyst is added to the system - no change

Water is removed from the system - shift to the right

Explanation:

When a constraint such as a change in temperature, pressure or volume is imposed on a reaction system in equilibrium, the equilibrium position will shift in such a way as to annul the constraint.

When the volume of a reaction system is increased, the equilibrium position shifts in the direction in which there is the highest total volume. This is the left hand side.

Since the reaction is exothermic (heat is given out) when the reaction is cooled down, the forward reaction is favoured.

Adding of reactants shifts the equilibrium position to the right hand side hence when H2 is added, the equilibrium position shifts to the right.

Decreasing the pressure shifts the equilibrium position to the direction of higher total volume hence the equilibrium shifts to the left when pressure is decreased.

A catalyst has no effect on the equilibrium position. It increases the rate of forward and reverse reaction to the same extent hence the equilibrium position is unaffected.

Removal of water from the system increases the rate of forward reaction since a product is being removed from the reaction system.

Answer:

5–bromo–9–chlorodecane

Explanation:

To name the compound given above, the following must be obtained:

1. The longest continuous carbon chain. This gives the parent name of the compound.

2. The substituent group attached to the compound.

3. Position of the substituent group.

4. Combine the above to obtain the name.

Now, we shall determine the name of the compound as follow:

1. The longest continuous carbon chain is 10. Thus, the parent name of the compound is decane.

2. The substituent groups attached to the compound are:

I. Bromine (Br) => Bromo

II. Chlorine (Cl) => Chloro

3. The position of the substituent groups are:

I. Br => carbon 5

II. Cl => carbon 9

NOTE: numbering is done alphabetically.

4. Therefore, the name of the compound is:

5–bromo–9–chlorodecane

Answer:

A.

Explanation:

I chose this answer and it was correct ‍♀️

Answer:

solo I

Explanation:

Según esta teoría para que se produzca una reacción deben cumplirse tres condiciones: Las moléculas de los reactivos tienen que chocar entre sí. Estos choques deben de producirse con energía suficiente de forma que se puedan romper y formar enlaces químicos.

Answer:

4

Explanation:

because it has 3 carbons  and 6 hydrogen

hope this helps :)

Answer:

Explanation:

a

Answer:

[tex]C_3H_7O_3[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to infer that the empirical formula of mannitol contains carbon, hydrogen and oxygen, so that the first step is to calculate the moles of C and H contained in the CO2 and H2O, respectively, as the only sources of these two elements in the formula:

[tex]n_C=1.993gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2} =0.0453molC\\\\n_H=0.9519gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.106molH[/tex]

Next, we calculate the grams and moles of O by subtracting the mass of C and H from the mass of the sample:

[tex]m_O=1.375g-0.0453molC*\frac{12gC}{1molC}-0.106molH*\frac{1.01gH}{1molH}=0.724gO\\\\n_O=0.724gO*\frac{1molO}{16.0gO} =0.0453molO[/tex]

Finally, we divide the moles of C, H and O by 0.0453 as the fewest moles of both C and O to find the mole ratios in the formula:

[tex]C:\frac{0.0453mol}{0.0453mol} =1\\\\H:\frac{0.106mol}{0.0453mol} =2.34\\\\O:\frac{0.0453mol}{0.0453mol} =1[/tex]

To get:

[tex]CH_{2.34}O[/tex]

Which must be multiplied by 3 to get whole numbers for all the subscripts, and therefore obtain:

[tex]C_3H_7O_3[/tex]

Regards!

Answer:

16974J of energy are required

Explanation:

The energy required is:

* The energy to heat solid water from -15°C to 0°C using:

q = m*S*ΔT

* The energy to convert the solid water to liquid water:

q = dH*m

* The energy required to increase the temperature of liquid water from 0°C to 75°C

q = m*S*ΔT

The first energy is:

q = m*S*ΔT

m = Mass water = 25g

S is specific heat of ice = 2.03J/g°C

ΔT is change in temperature = 0°C - (-15°C) = 15°C

q = 25g*2.03J/g°C*15°C

q = 761.3J

The second energy is:

q = dH*m

m = Mass water = 25g

dH is heat of fusion of water = 80cal/g

q = 80cal/g*25g

q = 2000cal * (4.184J/1cal) = 8368J

The third energy is:

q = m*S*ΔT

m = Mass water = 25g

S is specific heat of water= 4.184J/g°C

ΔT is change in temperature = 75°C-0°C = 75°C

q = 25g*4.184J/g°C*75°C

q = 7845J

The energy is: 7845J + 8368J + 761J =

16974J of energy are required