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alex41 [277]
3 years ago
8

One mole of silver has a mass of 107.9 grams. Approximately how many atoms of silver are present in one mole of silver

Chemistry
2 answers:
Katen [24]3 years ago
8 0

Answer: Silver has an atomic mass of aproximately 107.9.

Explanation:

1 mole of silver atoms = 107.9g and there are 6.02x10^23 atoms in one mole.

MAVERICK [17]3 years ago
3 0

Answer:

6.022 × 10²³ atoms Ag

Explanation:

Silver has a molar mass of 107.87. Convert grams to moles by dividing by the molar mass.

107. 9 grams × \frac{1 mole Ag}{107.87 grams} = 1 mole Ag

Multiply by Avogrado's number (6.022 × 10²³ atoms) to find the number of atoms.

1 mole <em>Ag × </em>6.022 × 10²³ atoms<em> </em>= 6.022 × 10²³ atoms Ag

There is approximately 6.022 × 10²³ atoms of silver present in one mole of silver.

Hope that helps.

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Answer:

B

Explanation:

I hope it helps you good luck

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3 years ago
Methylene chloride (CH2Cl2) has fewer chlorine atoms than chloroform (CHCl3). Nevertheless, methylene chloride has a larger mole
Talja [164]

Answer:

Explanation has been given below.

Explanation:

  • Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
  • Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
  • First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
  • Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
  • Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
  • Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.
6 0
3 years ago
Using the periodic table, choose the more reactive nonmetal.<br> Br or As
raketka [301]

Reactivity of non-metals depend on their ability to gain electrons. So, smaller is the size of a non-metal more readily it will attract electrons because then nucleus will be more closer to valence shell. ... Hence, Br is the non-metal which will be more reactive than At.

8 0
3 years ago
Read 2 more answers
Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir
guajiro [1.7K]

Answer:

y_{O2} =4.3%

Explanation:

The ethanol combustion reaction is:

C_{2}H_{5} OH+3O_{2}→2CO_{2}+3H_{2}O

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}

Dividing the previous equation by x:

1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles

Calculate the number of moles of CO2 and water considering the same:

0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)

0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)

The total number of moles at the reactor output would be:

N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)

So, the oxygen mole fraction would be:

y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3%

6 0
3 years ago
What type of reaction is: br2 + nal2 ---&gt; nabr + l2
Furkat [3]

Answer: its a single replacement reaction

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