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2 years ago

If a 100.-g sample of platinum metal has a volume of 4.67 ml, what is the density of platinum in g/cm3?

1 answer:

7 0

The density of any substance does not change at a certain temperature and pressure. Even though mass and volume are intensive properties (depends on the amount of substance), density is not. It is merely a fixed ratio of mass to volume. Therefore, the solution is

Density = Mass/Volume
For your information, quantitatively, cm³ is equivalent to mL.
Density = 100 g/4.67 cm³ = 21.41 g/cm³

You might be interested in

If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________. If the initial concen

Nikolay [14]

This is an incomplete question, here is a complete question.

If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________.

The reaction

$2NOBr(g)\rightarrow 2NO(g)+Br_2(g)$

It is a second-order reaction with a rate constant of 0.80 M⁻¹s⁻¹ at 11 °C.

Answer : The concentration of after 9.0 seconds is, 0.00734 M

Explanation :

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = 0.80 M⁻¹s⁻¹

t = time taken = 142 second

[A] = concentration of substance after time 't' = ?

$[A]_o$ = Initial concentration = 0.0440 M

Putting values in above equation, we get:

$0.80M^{-1}s^{-1}=\frac{1}{142s}\left (\frac{1}{[A]}-\frac{1}{(0.0440M)}\right)$

$[A]=0.00734M$

Hence, the concentration of after 9.0 seconds is, 0.00734 M

5 0

2 years ago

Given that cao(s) + h2o(l) → ca(oh)2(s), δh°rxn = –64.8 kj/mol, how many grams of cao must react in order to liberate 525 kj of

USPshnik [31]

Answer:

454.3 g.

Explanation:

From the given data:

1.0 mol of CaO liberates → – 64.8 kJ.

??? mol of CaO liberates → - 525 kJ.

∴ The no. of moles needed = (1.0 mol)(- 525 kJ)/(- 64.8 kJ) = 8.1 mol.

∴ The no. of grams of CaO needed = no. of moles x molar mass = (8.1 mol)(56.077 g/mol) = 454.3 g.

8 0

2 years ago

Calculate the wavelength of the photon emitted when an electron makes a transition from n=6 to n=3. You can make use of the foll

Angelina_Jolie [31]

Answer: The wavelength of light is $1.094\times 10^{-6}m$

Explanation:

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.097\times 10^7m^{-1}$

$n_f$ = Final energy level = 3

$n_i$ = Initial energy level = 6

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{3^2}-\frac{1}{6^2} \right )\\\\\lambda =\frac{1}{914617m^{-1}}=1.094\times 10^{-6}m$

Hence, the wavelength of light is $1.094\times 10^{-6}m$

6 0

2 years ago

If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is

mojhsa [17]

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a

compound.

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

Mn O

% composition 72.1 27.9

Divide by mass number 54.94 16

1.31 1.74

Divide by the smallest number 1.31 1.31

1 1.3

The resulting ratio is 1:1

Hence the Empirical formula is MnO, Manganese oxide

8 0

2 years ago

Consider the reaction: 2 Al + 3Br2 → 2 AlBr3 Suppose a reaction vessel initially contains 5.0 mole Al and 6.0 mole Br2. What is

timama [110]

Answer:

4 moles of $AlBr_3$ and 1 mole of $Al$ will be present in the reaction vessel

Explanation:

The reaction given in the question is

$2Al + 3 Br_2$ ⇒ $2AlBr_3$

According to the stoichiometric coefficients of the reaction, 2 moles of $Al$ requires 3 moles of $Br_2$ so in this reaction, $Br_2$ is a limiting reagent. So we will consider that $Al$ is in excess.

Now,

Since 3 moles of $Br_2$ requires 2 moles of $Al$

So, for 6 moles of $Br_2$ the moles of $Al$ required = $\frac{2}{3} \times 6$ = 4 moles.

Moles of $Al$ remaining after the completion of reaction = 5 - 4 = 1 mole.

Again,

Since 3 moles of $Br_2$ produces 2 moles of $AlBr_3$

So, moles of $AlBr_3$ produced by 6 moles of $Br_2$ = $\frac{2}{3} \times 6$ = 4 moles.

Therefore, after the completion of reaction, 4 moles of $AlBr_3$ and 1 mole of $Al$ will be present in the reaction vessel.

5 0

2 years ago