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3 years ago

Calculate the number of moles in 7.41 X 1017 atoms Ag.

Chemistry

1 answer:

jolli1 [7]3 years ago

4 0

Answer:

1.23x10^-6 mole

Explanation:

A clear understanding of Avogadro's hypothesis proved that 1mole of any substance contains 6.02x10^23 atoms. This indicates that 1mole of Ag contains 6.02x10^23 atoms.

Now, 1f 1mole of Ag contains 6.02x10^23 atoms, then Xmol of Ag will contain 7.41x10^17 atoms i.e

Xmol of Ag = 7.41x10^17/6.02x10^23 = 1.23x10^-6 mole

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Which is a balanced chemical equation? A. C7H16 + 5O2 6CO2 + 4H2O B. C7H16 + 11O2 7CO2 + 8H2O C. C7H16 + 14O2 7CO2 +5H2O D. C7H1

nadya68 [22]

Answer:

$\boxed{\text{B.}}$

Explanation:

$\rm C$_7$H_{16}$ + 11O$_2$ $\longrightarrow \,$ 7CO$_2$ + 8H$_2$O$

BALANCED. 7C, 16H, and 22O on each side of equation.

$\rm C$_7$H$_{16}$ + 5O$_2$ $\longrightarrow \,\rm 6CO$_2$ + 4H$_2$O$

NOT BALANCED. 7C on left and 6C on right.

$\rm C$_7$H$_{16}$ + 14O$_2$ $\longrightarrow \,$ 7CO$_2$ + 5H$_2$O$

NOT BALANCED. 16H on left and 10H on right.

$\rm C$_7$H$_{16}$ + 22O$_2$ $\longrightarrow \, $ 14CO$_2$ + 16H$_2$O$

NOT BALANCED. 7C on left and 14C on right.

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3 years ago

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2 years ago

A bond formed by electrical attraction between two oppositely charged ions is called

Komok [63]

Ionic bond, also called electrovalent bond , type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound.

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3 years ago

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}