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3 years ago

How many atoms are in 1.7 mols of CHF3

Chemistry

1 answer:

erastovalidia [21]3 years ago

5 0

Given :

Number of moles of CHF₃ is 1.7 .

Solution :

We know, 1 mole of any complex contains 6.022 × 10²³ molecules.

Let, 1.7 moles of CHF₃ contains n numbers of molecules.

So, n = 1.7 × 6.022 × 10²³ molecules

n = 10.2374 × 10²³ molecules

n = 1.0237 × 10²³ molecules

Hence, this is the required solution.

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What is the mass of 9.45 mol of aluminum oxide (Al2O3)?

Wittaler [7]

Answer:

mass = 963.53 grams

Explanation:

mass = moles x molar mass

m = 9.45 x 101.96 (molar mass of Al2O3)

m = 963.53g

3 0

2 years ago

Read 2 more answers

During lab, a student used a Mohr pipet to add the following solutions into a 25 mL volumetric flask. They calculated the final

kompoz [17]

Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃ = 0.200 M

V (Fe(NO₃)₃ = 10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

M (KSCN) = 0.00200 M

V (KSCN) = 1.42 mL

n (KSCN) = 0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃ + V (KSCN)

= 10.63 + 1.42

= 12.05 mL

Limiting reactant = KSCN

So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

= 0.000236 M

Excess reactant = (Fe(NO₃)₃

n(Fe(NO₃)₃ = 2.126 mmol - 0.00284 mmol

=2.123 mmol

For standard 2:

n (FeSCN⁺²) = 0.000236 * 4.63

=0.00109

V(standard 2) = 4.63 + 5.17

= 9.8 mL

M (FeSCN⁺²) = 0.00109/9.8

= 0.000111 M = 0.11 mM

Therefore, (FeSCN⁺²) = 0.11 mM

7 0

3 years ago

a water sample is found to have a cl- content of 100ppm as nacl what is the concentration of chloride in moles per liter

ladessa [460]

Answer:

The concentration of chloride ion is $2.82\times10^{-3}\;mol/L$

Explanation:

We know that 1 ppm is equal to 1 mg/L.

So, the $Cl^-$ content 100 ppm suggests the presence of 100 mg of $Cl^-$ in 1 L of solution.

The molar mass of $Cl^-$ is equal to the molar mass of Cl atom as the mass of the excess electron in $Cl^-$ is negligible as compared to the mass of Cl atom.

So, the molar mass of $Cl^-$ is 35.453 g/mol.

Number of moles = (Mass)/(Molar mass)

Hence, the number of moles (N) of $Cl^-$ present in 100 mg (0.100 g) of $Cl^-$ is calculated as shown below:

$N=\frac{0.100\;g}{35.453\;g/mol}=2.82\times 10^{-3}\;mol$

So, there is $2.82\times10^{-3}\;mol$ of $Cl^-$ present in 1 L of solution.

5 0

4 years ago

A ____ is the condensed, visible form of chromatin

Gelneren [198K]

a chromosomes is the condensed visible form of chromatin

6 0

3 years ago

Calculate the Equilibrium constant Kc at 25 C from the free - energy change for the following reaction . Zn(s) + 2Ag+(aq)<---

FinnZ [79.3K]

Answer: $Kc=1.24*10^5^2$

Explanation: For the given reaction:

$Kc=\frac{[Zn^+^2]}{[Ag^+]^2}$

Concentrations of the ions are not given so we need to think about another way to calculate Kc.

We can calculate the free energy change using the standard cell potential as:

$\Delta G^0=-nFE^0_c_e_l_l$

$E^0_c_e_l_l$ can be calculated using standard reduction potentials.

Standard reduction potential for zinc is -0.76 V and for silver, it is +0.78 V.

$E^0_c_e_l_l$ = $E^0_c_a_t_h_o_d_e+E^0_a_n_o_d_e$

Reduction takes place at anode and oxidation at cathode. As silver is reduced, it is cathode. Zinc is oxidized and so it is anode.

$E^0_c_e_l_l$ = 0.78 V - (-0.76 V)

$E^0_c_e_l_l$ = 0.78 V + 0.76 V

$E^0_c_e_l_l$ = 1.54 V

Value of n is two as two moles of electrons are transferred in the cell reaction F is Faraday constant and its value is 96485 C/mol of electron .

$\Delta G^0=-(2*96485*1.54)$

$\Delta G^0$ = -297173.8 J

Now we can calculate Kc using the formula:

$\Delta G^0=-RTlnKc$

T = 25+273 = 298 K

R = $8.314JK^-^1mol^-^1$

--297173.8 = -(8.314*298)lnKc

297173.8 = 2477.572*lnKc

$lnKc=\frac{297173.8}{2477.572}$

lnKc = 119.946

$Kc=e^1^1^9^.^9^4^6$

$Kc=1.24*10^5^2$

5 0

3 years ago