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Alisiya [41]
3 years ago
15

How is an electrolytic cell different from a galvanic cell?

Chemistry
1 answer:
LenKa [72]3 years ago
4 0

Answer:

<em>b</em>

Explanation:

<em>a p e x :)</em>

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How many moles of solute particles are present in 1 ml of aqueous 0.020 m (nh4)2co3?
Dima020 [189]
Vs = 1.0 mL = 0.001 L
c((NH4)2CO3) = <span>0.02 M
n(</span>(NH4)2CO3) = ?

For the purpose, here we will use the next equation:

c=n/V ⇒ n=cxV

n((NH4)2CO3) = 0.02M x 0.001L 

n((NH4)2CO3) = 2x10⁻⁵ mole of (NH4)2CO3 is presented in the solution


6 0
3 years ago
Which of the following statement is correct for a group of a periodic table.
Vaselesa [24]

Answer:

I believe it is B

Explanation:

I am on this lesson too :)

6 0
3 years ago
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How would doubling the concentration of crystal violet affect the reaction rate?
Helga [31]

Actually the rate of reaction can be shown as:

R = k [A]^x [B]^y

The rate of reaction can be determined with help of order of reaction which is an experimental quantity and the order can be determined experimentally.

Hence for the rate of reaction crystall violet depends on order of crystall violet and mostly the order of crystall violet is 1 and thus on doubling the concentration, the rate of reaction gets doubles.

5 0
3 years ago
URGENT!!!
Inga [223]

Answer:

{ \bf{from \: ionic \: product \: of \: water : }} \\ {  \boxed{ \tt{k _{w} = [H _{3} O {}^{ + } ][OH {}^{ - } ]}}} \\  \\ { \tt{1 \times  {10}^{ - 14} = (1 \times  {10}^{ - 5} ) [OH {}^{ - } ]}} \\  \\ { \tt{[OH {}^{ - } ] = 1 \times  {10}^{ - 9} }} \: M

6 0
3 years ago
A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) C
Mamont248 [21]

Answer:

T_{eq}=23.85^oC

Explanation:

Hello,

In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

\Delta H_{Cu}=-\Delta H_{H_2O}

Therefore the equilibrium temperature shows up as:

m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC

Best regards.

6 0
3 years ago
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