A species that changes the rate of a reaction but is neither consumed nor changed is

Explain how the equation for GPE = force x distance.

Mkey [24]

GPE is the gravitational Potential Energy which is given by mgh.

mg is the gravitational force(a force) and h is the height of the object(or distance of the object from earth's surface)

SO GPE = mg (force) x h (distance)

4 0

3 years ago

171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in o

Fantom [35]

The complete question is as follows: 171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in one liter of water at 25 oC. At 25 oC the vapor pressure of water is 24 mmHg. Which value is closest to the vapor pressure (VP) of this solution at 25 oC?

a. 16mm Hg

b. 24mm Hg

c. 20mm Hg

d. 12mm Hg

Answer: The vapor pressure (VP) of this solution at $25^{o}C$ is closest to the value 24 mm Hg.

Explanation:

Given: Mass of sucrose = 171 g

Mass of water = 1 L = 1000 g

Vapor pressure of water = 24 mm Hg

As moles is the mass of substance divided by its molar mass. Hence, moles of water (molar mass = 18.02 g) is calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{1000 g}{18.02 g/mol}\\= 55.49 mol$

Similarly, moles of sucrose (molar mass = 342 g/mol) is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{171 g}{342 g/mol}\\= 0.5 mol$

Total moles = 55.49 + 0.5 mol = 55.99 mol

Mole fraction of water is as follows.

$Mole fraction = \frac{moles of water}{total moles}\\= \frac{55.49}{55.99}\\= 0.99$

Formula used to calculate vapor pressure of the solution is as follows.

$P_{i} = P^{o}_{i} \times \chi_{i}$

where,

$P_{i}$ = vapor pressure of component i over the solution

$P^{o}_{i}$ = vapor pressure of pure component i

$\chi_{i}$ = mole fraction of i

Substitute the values into above formula to calculate vapor pressure of water as follows.

$P_{i} = P^{o}_{i} \times \chi_{i}\\= 24 mm Hg \times 0.99\\= 23.76 \\or 24 mm Hg\\$

Thus, we can conclude that the vapor pressure (VP) of this solution at $25^{o}C$ is closest to the value 24 mm Hg.

4 0

3 years ago

A 0.8115 g sample of HCl was placed into a 50 mL volumetric flask and the sample was thoroughly dissolved in water to make 50 mL

inysia [295]

Answer:

Molarity of NaOH solution is 1.009 M

Explanation:

Molar mass of HCl is 36.46 g/mol

Number moles = (mass)/(molar mass)

So, 0.8115 g of HCl = $\frac{0.8115}{36.46}moles$ HCl = 0.02226 moles HCl

1 mol of NaOH neutralizes 1 mol of HCl.

So, if molarity of NaOH solution is S(M) then moles of NaOH required to reach endpoint is $\frac{S\times 22.07}{1000}moles$

So, $\frac{S\times 22.07}{1000}=0.02226$

or, S = 1.009

So, molarity of NaOH solution is 1.009 M

5 0

3 years ago

Which of the following is not true of alloys?

Scilla [17]

Option d: copper.

Because copper is an element, not a mixture.

7 0

3 years ago

A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an

erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

7 0

3 years ago