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shusha [124]
2 years ago
8

According to the VSEPR model, the arrangement of electron pairs around NH3 and CH4 is A. the same, because in each case there ar

e the same number of electron pairs around the central atom B. different or the same, depending on the conditions leading to maximum repulsion C. different, because in each case there are a different number of atoms around the central atom D. the same, because both nitrogen and carbon are both in the second period E. different, because in each case there are a different number of electron pairs around the central atom
Chemistry
1 answer:
aleksklad [387]2 years ago
3 0

Answer:

Option "B" is correct.

Explanation:

According to VSEPR theory, There are repulsion forces exists among the bond pair - bond pair or bond pair - lone pair of electrons. In NH_{3} and  CH_{4}, the number of electron pairs are same but methane has all the four bond pairs where in ammonia, three bond pairs and one lone pair exists. And thus there are repulsion forces possible in between the lone pair and bond pair of electrons thus the arrangement of electron pairs around both the molecules is same or different depending up on the conditions leading to maximum repulsion.  

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Which solution contains the largest number of moles of chloride ions?
elixir [45]
Molarity = number of moles of solute/liters of solution
number of moles of solute = molarity x liters of solution

Part (a): <span>30.00 ml of 0.100m Cacl2
number of moles of CaCl2 =  0.1 x 0.03 = 3x10^-3 moles
1 mole of CaCl2 contains 2 moles of chlorine, therefore 3x10^-3 moles of CaCl2 contains 6x10^-3 moles of chlorine

Part (b): </span><span>10.0 ml of 0.500m bacl2
number of moles of BaCl2 = 0.5 x 0.01 = 5x10^-3 moles
1 mole of BaCl2 contains 2 moles of chlorine, therefore 5x10^-3 moles of BaCl2 contains 10x10^-3 moles of chlorine

Part (c): </span><span>4.00 ml of 1.000m nacl
number of moles of NaCl = 1 x 0.004 = 0.004 moles
1 mole of NaCl contains 1 mole of chlorine, therefore 4x10^-3 moles of NaCl contains 4x10^-3 moles of chlorine

Part (d): </span><span>7.50 ml of 0.500m fecl3
number of moles of FeCl3 = 0.5 x 0.0075 = 3.75x10^-3 moles
1 mole of FeCl3 contains 3 moles of chlorine, therefore 3.75x10^-3 moles of FeCl3 contains 0.01125 moles of chlorine

Based on the above calculations, the correct answer is (d)</span>
5 0
2 years ago
Read 2 more answers
Which of the following compounds would be most effective in lowering the melting point of ice on roads? a) CaCl2 b) NaCl C) K3PO
Mrac [35]
I'm not 100% sure on this, but I would go with C) NaCl. 
NaCl is a salt, and that is used to melt the ice on the roads. Hope this helps!
7 0
2 years ago
You have a 16.0-oz. (473-mL) glass of lemonade with a concentration of 2.66 M. The lemonade sits out on your counter for a coupl
salantis [7]

<u>Answer:</u> The new concentration of lemonade is 3.90 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}     .....(1)

Molarity of lemonade solution = 2.66 M

Volume of solution = 473 mL

Putting values in equation 1, we get:

2.66M=\frac{\text{Moles of lemonade}\times 1000}{473}\\\\\text{Moles of lemonade}=\frac{2.66\times 473}{1000}=1.26mol

Now, calculating the new concentration of lemonade by using equation 1:

Moles of lemonade = 1.26 moles

Volume of solution = (473 - 150) mL = 323 mL

Putting values in equation 1, we get:

\text{New concentration of lemonade}=\frac{1.26\times 1000}{323}\\\\\text{New concentration of lemonade}=3.90M

Hence, the new concentration of lemonade is 3.90 M

7 0
1 year ago
consideras util conocer las propiedades extensivas e intensivas de los insumos utilizados para la elaboración de producto ¿por q
Brums [2.3K]

Answer:

Explanation:

No.

Las propiedades físicas de los materiales y sistemas a menudo se pueden clasificar como intensivas o extensivas, según cómo cambia la propiedad cuando cambia el tamaño (o extensión) del sistema. Según la IUPAC, una cantidad intensiva es aquella cuya magnitud es independiente del tamaño del sistema, mientras que una cantidad extensiva es aquella cuya magnitud es aditiva para los subsistemas. Esto refleja las ideas matemáticas correspondientes de media y medida, respectivamente.

Una propiedad intensiva es una propiedad a granel, lo que significa que es una propiedad física local de un sistema que no depende del tamaño del sistema o de la cantidad de material en el sistema. Los ejemplos de propiedades intensivas incluyen temperatura, T; índice de refracción, n; densidad, ρ; y dureza de un objeto.

Por el contrario, propiedades extensivas como la masa, el volumen y la entropía de los sistemas son aditivas para los subsistemas porque aumentan y disminuyen a medida que crecen y se reducen, respectivamente.  

Estas dos categorías no son exhaustivas, ya que algunas propiedades, físicas no son exclusivamente intensivas ni extensivas. Por ejemplo, la impedancia eléctrica de dos subsistemas es aditiva cuando, y solo cuando, se combinan en serie; mientras que si se combinan en paralelo, la impedancia resultante es menor que la de cualquiera de los subsistemas.

¡Espero haberte ayudado!  :)

7 0
1 year ago
Q1) A vapor-compression refrigeration system operates on the cycle of Fig. 9.1. The refrigerant is 1,1,1,2-Tetrafluoroethane. Gi
hoa [83]

Answer:

i)   0.5071 (kg/s)

ii)  -1407.1 kj/kg

iii)  204.05 Kw

iv)  5.881

v)    9.238

Explanation:

Given Data:

evaporation temperature ( T ) = 4°c = 277.15 K

Condensation Temperature ( T ) = 34°c = 307.15 K

<em>n</em> ( compressor efficiency ) = 0.76

refrigeration rate = 1200 kJ.s^-1

i) determine the circulation rate of the refrigerant

m = \frac{Q}{H2 - H1}  = \frac{Q}{H2 - H4\\}  ------- 1

Q = 1200 Kj.s^-1

H2 = entropy at step 2 = 2508.9 (kJ / kg ) ( gotten from Table F )

H4 = entropy at step 4 = 142.4 ( kJ/ kg )

back to equation 1

m ( circulation rate of refrigerant ) = 0.5071 (kg/s)

ii) heat transfer rate in the condenser

Q = m ( H4 - H3 )

    = 0.5071 ( 142.4 - 2911.27 )

    = -1407.1 kj/kg

where H3 = H2 + ΔH23 = 2911.27 (kj/kg) ( as calculated )

iii) power requirement

w = m * ΔH23

   = 0.5071 (kg/s) * 402.37 (kj/kg) =  204.05 Kw

where: ΔH23 = \frac{H'3 - H2 }{0.76} = \frac{2814.7-2508.9}{0.76} = 402.37 (kj/kg)

iv) coefficient of performance of a cycle

W = Qc / w

  = 1200 Kj.s^-1/ 204.05 kw

  = 5.881

v) coefficient of performance of a Carnot refrigeration cycle

w_{carnot} = \frac{T2}{T4 - T2}

            =  277.15 / ( 307.15 - 277.15 )

            = 9.238

4 0
1 year ago
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