A) How many moles of gold atoms are present in a 295-gram sample of gold? B) How many gold atoms would there be in the sample?

Answers

Answer 1
Answer: The molar mass of gold, Au, as seen on the periodic table, is 196.97 grams per mole. So, do this to find the number of moles in a 295 gram sample:

(295g\ Au)((1mol\ Au)/(196.97g\ Au)) = 1.49769mol\ Au

So, a 295 gram sample of gold has about 1.5 moles of gold in it.

To find the number of atoms, multiply the number of moles by Avogadro's number:

(1.49769mol\ Au)((6.022*10^(23)\ atoms)/(mol)) = 9.02 * 10^(23)\ atoms\ Au
Answer 2
Answer: 5 gram of 197 Au contain 1.53*10^22 atoms. so just multiply that with 59 to get 295 grams :)

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Which of the following is a measure of randomness in a system?A) entropy.
B) kinetic energy.
C) potential energy.
D) chemical energy.

Answers

Answer:

A) entropy

Explanation:

The atomic number and mass number do not change during this type of radioactive decay.

Answers

In gamma decay, no change in proton number occurs, so the atom does not become a different element

Which isotope is used to date ancient artifacts such as fossils

Answers

Answer: Carbon-14 isotope is used to date ancient artifacts such as fossils.

Explanation:

To determine the age of ancient artifacts such as fossils, we use radiocarbon dating process.

In this process, a naturally occurring radioisotope of carbon which is Carbon-14 is used. This isotope is generally used because this isotope gets constantly created in the atmosphere with the interaction of cosmic rays with the nitrogen present in atmosphere.

Hence, an isotope of carbon which is Carbon-14 is used to determine the age of ancient artifacts such as fossils.

dating:
this dating is called CARBON 14 DATING...because Carbon 14 is the isotope used to date fossils...
Also Potassium 40 is also used in some cases....

A closed system initially containing 1×10^-3M H2 and 2×10^-3 at 448°C is allowed to reach equilibrium. Analysis of the equilibrium mixture show s that the concentration of HI is 1.87×10^ -3. Calculate Kc at 448°C for the reaction.

Answers

Answer:

1.74845

Explanation:

We have the following reaction:

I2 + H2 => 2 HI

Now, the constant Kc, has the following formula:

Kc = [C] ^ c * [D] ^ d / [A] ^ a * [B] ^ b

In this case I2 is A, H2 is B and C is HI

We know that the values are:

 H2 = 1 × 10 ^ -3 at 448 ° C

I2 = 2 × 10 ^ -3 at 448 ° C

HI = 1.87 × 10 ^ -3 at 448 ° C

Replacing:

Kc = [1.87 × 10 ^ -3] ^ 2 / {[2 × 10 ^ -3] ^ 1 * [1 × 10 ^ -3] ^ 1}

Kc = 1.87 ^ 2/2 * 1

Kc = 1.74845

Which means that at 448 ° C, Kc is equal to 1.74845

Answer:

K_c = 51

Explanation:

[H2] = 10^-3

[I2] = 2*10^-3

[HI] = 0

in equilbiirum

[H2] = 10^-3 - x

[I2] = 2*10^-3 -x

[HI] = 0 + 2x

and we know

[HI] = 0 + 2x = 1.87*10^-3

x = ( 1.87*10^-3)/2 =  0.000935

then

[H2] = 10^-3 - 0.000935 = 0.000065

[I2] = 2*10^-3 -0.000935 = 0.001065

                        H₂                +          I              ⇄              2 HI

Initially     1 × 10⁻³                       2 × 10⁻³

Change  -9.35 × 10⁻⁴               -9.35 × 10⁻⁴                +1.87 × 10⁻³

At equil   6.5 × 10⁻⁵                  1.06 5 × 10⁻³               1.87 × 10⁻³

HI increase by 1.87 × 10⁻³M

K_c = ([HI]^2)/([H_2][I_2]) \n\n= ((1.87*10^-^3)^2)/((6.5*10^-^5)(1.065*10^-3)) \n\nK_c = 51

Complete the following Empirical Formula & Molecular Formula problem:2.) A compound containing C and Cl has a molar mass of 284.77 g/mol and an empirical formula of CCl.

Answers

Empirical formula is CCl.

Since the molar mass is provided, it is safe to assume that the empirical formula is not equal to the molecular formula.

We need to get the empirical weight to determine the multiple needed to arrive to the correct molecular formula.

Atom          Number in Molecule            Atomic Weight        Total Mass
C                              1                            12.0107                  12.0107
Cl                              1                            35.453                    35.453
                                                          total empirical weight    47.4637

Molecular weight / Empirical Weight
284.77 g/mol / 47.46 g/mol = 6 - the multiple needed to be multiplied with the empirical formula to get the molecular formula.

C has a molecular number of 1 =>  1 * 6 = 6  Thus C₆
Cl has a molecular number of 1 => 1 * 6 = 6 Thus Cl₆

Molecular Formula is C₆CL₆

A gas occupies 12.3 l at a pressure of 40 mm hg. what is the volume when the pressure is increased to 60 mm hg

Answers

From boyles law.
P1v1=P2v2
So,12.3×40=v×60
V= 12.3×40/60
8.2
Boyle's Law states an inverse relationship between volume and pressure. If the constant of proportionality is 12.3×40 which is 492, the volume would be 492÷60 which is 8.2.

Answer is 8.2 I