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3 years ago

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Chemistry

2 answers:

Naily [24]3 years ago

8 0

Answer:

D) He did not multiply the chlorine and oxygen atoms by the coefficient 4.

Explanation:

The coefficient 4 at the beginning of the chemical formula indicates that there are four Ca(ClO3)2 molecules. Think of this as Ca(ClO3)2 × 4. This means that he had to multiply the number of atoms for each element by 4 as well, so he should've ended up with 4 total calcium atoms (which is correct), 8 total chlorine atoms, and and 24 total oxygen atoms. He did not get all these answers because he didn't multiply the chlorine and oxygen atoms by the coefficient 4.

NARA [144]3 years ago

4 0

Answer:

D) He did not multiply the chlorine and oxygen atoms by the coefficient 4.

Explanation:

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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

4 years ago

the temperature of a sample of copper increased by 23.0 C when 265 J of heat was applied. What is the mass of the sample?

Anuta_ua [19.1K]

Answer: 29.93g

Explanation:

Q = 265J

C = 0.385J/g/°C

Δt = 23°C

M =?

Q = MCΔT

M = Q / CΔT = 265 / (0.385x23)

M = 29.93g

3 0

4 years ago

A mixture of helium and nitrogen gases, in a 7.03 L flask at 17 °C, contains 0.738 grams of helium and 8.98 grams of nitrogen. T

Andrei [34K]

Answer:

The partial pressure of nitrogen in the flask is 1.08 atm and the total pressure in the flask is 1.70 atm.

Explanation:

We must use the Ideal Gas Law to solve this:

Pressure . volume = n . R . T

T = T° in K → T°C + 273

17°C + 273 = 290K

n = moles

In a mixture, n is the total moles (Sum of each mol, from each gas)

Moles = Mass / Molar mass

Moles He = 0.738 g / 4g/m = 0.184 moles

Moles N₂ = 8.98 g / 28g/m = 0.320 moles

0.184 m + 0.320m = 0.504 moles

P . 7.03L = 0.504m . 0.082L.atm/ mol.K . 290K

P = (0.504m . 0.082L.atm/ mol.K . 290K) /7.03L

P = 1.70 atm - This is the total pressure.

To know the partial pressure of N₂ we can apply, the molar fraction:

Moles of N₂ / Total moles = Pressure N₂ / Total pressure

0.320m / 0.504m = Pressure N₂ / 1.70atm

(0.320m / 0.504m) . 1.70atm = Pressure N₂

1.08atm = Pressure N₂

8 0

4 years ago

Please help me..........

Kobotan [32]

Answer:

4/6

6/9

8/12

Explanation:

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5 0

3 years ago

1. When a metal and a(n) ___ bond, they do not share electrons. 2. The ___ atom transfers one or more valence electrons to the _

NISA [10]

Answer:

1 . Nonmetal

2. metal, non metal

3. Ion

4. positive

5 0

4 years ago