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At a given temperature, the equilibrium constant for the formation of HI from H2 and I2 was found to be 29.9. Calculate the equi

evablogger [386]

Answer:

The correct answer is: K'= 0.033.

Explanation:

The formation of HI from H₂ and I₂ is given by:

H₂ + I₂ → 2 HI K= 29.9

The decomposition of HI is the reverse reaction of the formation of HI:

2 HI → H₂ + I₂ K'

Thus, K' is the equilibrium constant for the reverse reaction of formation of HI. It is calculated as the reciprocal of the equilibrium constant of the forward reaction (K):

K' = 1/K = 1/(29.9)= 0.033

Therefore, the equilibrium constant for the decomposition of HI is K'= 0.033

5 0

4 years ago

Air thermals are updrafts of warm air that rise from

Anastaziya [24]

Answer:

B. convection because warm air rises and cool air sinks.

Explanation:

The processes that enables birds to fly to higher altitudes without flapping their winds while riding air thermals is called convection.

Convection is one of the forms of heat transfer usually found in gases and liquids.

In convection density differences due to temperature changes causes the circulation.
Warm air is less dense and it is hotter.
It rises up from the surface where it is warmed
Cold air sinks to replace the warm air. Cold air form when the air loses its energy to the surrounding.
By virtue of this, birds are able to take advantage of it for their flight.

7 0

3 years ago

Describe, using scientific terms, how plants turn sunlight into energy? make sure to refer to the chemical equation to photosynt

Sergio039 [100]

A plant is a living organism that uses chlorophyll in the chloroplasts to perform photosynthesis. The chemical equation of photosynthesis is
Sunlight(Photons)+ CO2+2H2O ----->C6H12O6+H2O+O2.(Not balanced) The reactants are carbon dioxide and water. The products are water and glucose with oxygen as a waste product.Photosynthesis can make a plant form a closed cycle as plants need to respire with oxygen which is a waste product of one of their reactions.

5 0

4 years ago

Read 2 more answers

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti

ella [17]

Answer :

The mass of excess mass of $Na_2CO_3$ , $AgNO_3,Ag_2CO_3\text{ and }NaNO_3$ are, 1.908 g, 0 g, 12.144 g and 3.74 g respectively.

Explanation : Given,

Mass of $Na_2CO_3$ = 4.25 g

Mass of $AgNO_3$ = 7.50 g

Molar mass of $Na_2CO_3$ = 106 g/mole

Molar mass of $AgNO_3$ = 170 g/mole

Molar mass of $Ag_2CO_3$ = 276 g/mole

Molar mass of $NaNO_3$ = 85 g/mole

First we have to calculate the moles of $Na_2CO_3$ and $AgNO_3$ .

$\text{Moles of }Na_2CO_3=\frac{\text{Mass of }Na_2CO_3}{\text{Molar mass of }Na_2CO_3}=\frac{4.25g}{106g/mole}=0.040moles$

$\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=\frac{7.50g}{170g/mole}=0.044moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Na_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2NaNO_3$

From the balanced reaction we conclude that

As, 2 moles of $AgNO_3$ react with 1 mole of $Na_2CO_3$

So, 0.044 moles of $AgNO_3$ react with $\frac{0.044}{2}=0.022$ moles of $Na_2CO_3$

From this we conclude that, $Na_2CO_3$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

The excess mole of $Na_2CO_3$ = 0.040 - 0.022 = 0.018 mole

Now we have to calculate the mass of excess mole of $Na_2CO_3$ .

$\text{Mass of }Na_2CO_3=\text{Moles of }Na_2CO_3\times \text{Molar mass of }Na_2CO_3=(0.018mole)\times (106g/mole)=1.908g$

Now we have to calculate the moles of $Ag_2CO_3$ .

As, 1 moles of $AgNO_3$ react to give 1 moles of $Ag_2CO_3$

So, 0.044 moles of $AgNO_3$ react to give 0.044 moles of $Ag_2CO_3$

Now we have to calculate the mass of $AgCO_3$ .

$\text{Mass of }Ag_2CO_3=\text{Moles of }Ag_2CO_3\times \text{Molar mass of }Ag_2CO_3=(0.044mole)\times (276g/mole)=12.144g$

Now we have to calculate the moles of $NaNO_3$ .

As, 2 moles of $AgNO_3$ react to give 2 moles of $NaNO_3$

So, 0.044 moles of $AgNO_3$ react to give 0.044 moles of $NaNO_3$

Now we have to calculate the mass of $NaNO_3$ .

$\text{Mass of }NaNO_3=\text{Moles of }NaNO_3\times \text{Molar mass of }NaNO_3=(0.044mole)\times (85g/mole)=3.74g$

6 0

4 years ago

How many moles are in a 12.0 g sample of NiC12

Nady [450]

Answer:

0.17 moles

Explanation:

In the elements of the periodic table, the atomic mass = molar mass. <u>Ex:</u> Atomic mass of Carbon is 12.01 amu which means molar mass of Carbon is also 12.01g/mol.

In order to find the # of moles in a 12 g sample of NiC-12, we will need to multiply the number of each atom by its molar mass and then add the masses of both Nickel and C-12 found in the periodic table:

Molar Mass of Ni (Nickel): 58.69 g/mol

Molar Mass of C (Carbon): 12.01 g/mol

Since there's just one atom of both Carbon and Nickel, we just add up the masses to find the molar mass of the whole compound of NiC-12.

58.69 g/mol of Nickel + 12.01 g/mol of Carbon = 70.7 g/mol of NiC-12

There's 12g of NiC-12, which is less than the molar mass of NiC-12, so the number of moles should be less than 1. In order to find the # of moles in NiC-12, we need to do some dimensional analysis:

12g NiC-12 (1 mol of NiC-12/70.7g NiC-12) = 0.17 mol of NiC-12
The grams cancel, leaving us with moles of NiC-12, so the answer is 0.17 moles of NiC-12 in a 12 g sample.

<em>P.S. C-12 or C12 just means that the Carbon atom has an atomic mass of 12amu and a molar mass of 12g/mol, or just regular carbon.</em>

5 0

3 years ago