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Effectus [21]
3 years ago
9

Tia has a sample of pure gold (Au). She weighed the sample and the result was 35.9 grams. Tia wants to determine the number of a

toms in the sample. Calculate the number of atoms in 35.9 g of pure gold.
Chemistry
2 answers:
Gemiola [76]3 years ago
4 0
First, find moles of gold given the mass of the sample:
(35.9g Au)/(197.0g/mol Au) = 0.182mol Au

Second, multiply moles of Au by Avogrado's number:
(0.182mol)(6.02 x10^23)= 1.10x10^23 atoms Au
IRINA_888 [86]3 years ago
3 0

Answer: 1.09\times 10^{23atoms of gold.

Explanation: To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}

For gold

Mass of gold given = 35.9 g

Molar mass of gold = 197 g/mol

Putting values in above equation, we get:

\text{Moles of gold}=\frac{35.9g}{197g/mol}=0.18mol

According to Avogadro's law, 1 mole of every substance contains avogadro's number (6.023\times 10^{23}) of particles.

Thus 0.18 moles of gold will contain=\frac{6.023\times 10^{23}}{1}\times 0.18=1.09\times 10^{23} atoms of gold.

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Assoli18 [71]

Answer:

m_{Co^{3+}}=0.563gCo^{3+}

Explanation:

Hello there!

In this case, since these mole-mass relationships are understood in terms of the moles of the atoms forming the considered compound, we first realize that the chemical formula of the cobalt (III) nitrate is Co(NO₃)₃ whereas there is a 1:1 mole ratio of the cobalt (III) ion (molar mass = 58.93 g/mol) to the entire compound. In such a way, we first compute the moles of the salt (molar mass = 58.93 g/mol) and then apply the aforementioned mole ratio to obtain the grams of the required cation:

m_{Co^{3+}}=2.34gCo(NO_3)_3*\frac{1molCo(NO_3)_3}{244.95 gCo(NO_3)_3} *\frac{1molCo^{3+}}{1molCo(NO_3)_3} *\frac{58.93gCo^{3+}}{1molCo^{3+}} \\\\m_{Co^{3+}}=0.563gCo^{3+}

Best regards!

4 0
2 years ago
In E. coli, the enzyme hexokinase catalyzes the reaction: Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Ke
kondor19780726 [428]

Answer:

Explanation:

Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.

In the living E. coli cells,

[ATP] = 7.9 mM;

[ADP] = 1.04 mM,

[glucose] = 2 mM,

[glucose 6-phosphate] = 1 mM.

Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.

The reaction is given as

Glucose + ATP → glucose 6-phosphate + ADP

Now reaction quotient for given equation above is

q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}

q=\frac{(1mm)\times (1.04 mm)}{(7.9mm)\times (2mm)} \\\\=6.582\times 10^{-2}

so,

q ⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable  until q = Keq

4 0
3 years ago
How many moles of helium are 8.84×10^24 atoms of He?
viva [34]

Answer:

14.77 mol.

Explanation:

  • It is known that every 1.0 mole of compound or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

<u><em>Using cross multiplication:</em></u>

1.0 mole of He contains → 6.022 x 10²³ atoms.

??? mole of He contains → 8.84 x 10²⁴ atoms.

<em>∴ The no. of moles of He contains (8.84 x 10²⁴ atoms) </em>= (1.0 mol)(8.84 x 10²⁴ atoms)/(6.022 x 10²³ atoms) =<em> 14.77 mol.</em>

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8 0
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sertanlavr [38]

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Explanation:

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