How does the sun transfer heat energy to various materials?

As the movement of molecules within a substance increases, describe the change in heat of the substance. Justify your response i

Vladimir [108]

It will get hotter, because the molecules create heat when they move around just like us when we run.

3 0

3 years ago

Read 2 more answers

How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

8 0

3 years ago

What type of model did Rutherford, Thompson and Bohr develop?

motikmotik

The Bohr atomic model, relying on quantum mechanics, built upon the Rutherford model to explain the orbits of electrons.

4 0

3 years ago

For the reaction

s2008m [1.1K]

Answer:

0.558mole of SO₃

Explanation:

Given parameters:

Molar mass of SO₃ = 80.0632g/mol

Mass of S = 17.9g

Molar mass of S = 32.065g/mol

Number of moles of O₂ = 0.157mole

Molar mass of O₂ = 31.9988g/mol

Unknown:

Maximum amount of SO₃

Solution

We need to write the proper reaction equation.

2S + 3O₂ → 2SO₃

We should bear in mind that the extent of this reaction relies on the reactant that is in short supply i.e limiting reagent. Here the limiting reagent is the Sulfur, S. The oxygen gas would be in excess since it is readily availbale.

So we simply compare the molar relationship between sulfur and product formed to solve the problem:

First, find the number of moles of Sulfur, S:

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

Number of moles of S = $\frac{17.9 }{32.065}$ = 0.558mole

Now to find the maximum amount of SO₃ formed, compare the moles of reactant to the product:

2 mole of Sulfur produced 2 mole of SO₃

Therefore; 0.558mole of sulfur will produce 0.558mole of SO₃

5 0

3 years ago

A buffer solution is composed of 7.50 mol of acid and 4.75 mol of the conjugate base. If the pKa of the acid is 4.90 , what is t

Mamont248 [21]

Answer:

The correct answer is: pH= 4.70

Explanation:

We use the <em>Henderson-Hasselbach equation</em> in order to calculate the pH of a buffer solution:

$pH= pKa + log \frac{ [conjugate base]}{[acid]}$

Given:

pKa= 4.90

[conjugate base]= 4.75 mol

[acid]= 7.50 mol

We calculate pH as follows:

pH = 4.90 + log (4.75 mol/7.50 mol) = 4.90 + (-0.20) = 4.70

7 0

3 years ago