What two spheres are being pulled by the fires of gravity

Rewrite each equation below with the delta h value included with either the reactants or he products , and identify the reaction

grin007 [14]

I will take a stab at it, but there are not equations, did you forget them?

5 0

3 years ago

A sample of 87.6 g of carbon is reacted with 136 g of

Vadim26 [7]

Answer:

A. fluorine, 1.79 moles

Explanation:

Given parameters:

Mass of carbon = 87.7g

Mass of fluorine gas = 136g

Unknown:

The limiting reactant and the maximum amount of moles of carbon tetrafluoride that can be produced = ?

Solution:

Equation of the reaction:

C + 2F₂ → CF₄

let us find the number of the moles the given species;

Number of moles = $\frac{mass}{molar mass}$

C; molar mass = 12;

Number of moles = $\frac{87.7}{12}$ = 7.31moles

F; molar mass = 2(19) = 38g/mol

Number of moles = $\frac{136}{38}$ = 3.58moles

So;

From the give reaction:

1 mole of C requires 2 moles of F₂

7.31 moles of C will then require 2 x 7.31 moles of F₂ = 14.62moles

But we have 3.58 moles of the F₂;

Therefore, the reactant in short supply is F₂ and it is the limiting reactant;

So;

2 moles of F₂ will produce mole of CF₄

3.58 moles of F₂ will then produce $\frac{3.58}{2}$ = 1.79moles of CF₄

6 0

3 years ago

a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh

Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

$Molarity = M = \frac{1,22 mol solute}{lts solution}$

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

$1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute$

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

$1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution$

With the molecular weight of solute (Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol), we can obtain the mass of solute:

$1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute$

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

$molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal$

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

$%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%$

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: Moles solution = moles solute + moles solvent. First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

$702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent$

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

$Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03$

6 0

3 years ago

What temperature air has low pressure?

finlep [7]

Explanation:

A low-pressure area, or "low", is a region where the atmospheric pressure at sea level is below that of surrounding locations. Low-pressure systems form under areas of wind divergence that occur in upper levels of the troposphere.

5 0

3 years ago

For a reaction at equilibrium, which change can increase the rates of the forward and reverse reactions?a decrease in the concen

Monica [59]

Question requires a change resulting in an increase in both forward and reverse reactions. Now lets discuss options one by one and see there impact on rate of reactions.

1) A decrease in the concentration of the reactants:
When concentration of reactant is decreased it will shift the equilibrium in Backward direction, so resulting in increasing the backward reaction and decreasing the forward direction. Hence, this option is incorrect.

2) A decrease in the surface area of the products:
Greater the surface Area greater is the chances of collision and greater will be the rate of reaction. As the surface area of products is decreased it will not favor the backward reaction. Hence again this statement is incorrect according to given statement.

3) An increase in the temperature of the system:
An increase in temperature will shift the reaction in endothermic side. Hence, if the reaction is endothermic, an increase in temperature will increase the rate of forward direction or if the reaction is exothermic it will increase the rate of reverse direction. Hence, this option is correct according to given statement.

4) An increase in the activation energy of the forward reaction:
An increase in Activation energy will decrease the rate of reaction, either it is forward or reverse. So this is incorrect.

Result:
Hence, the correct answer is,"An increase in the temperature of the system".

7 0

3 years ago