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2 years ago

Is the answer B? Help

Chemistry

1 answer:

Masja [62]2 years ago

7 0

Answer:

Explanation:

Hmm, so we have the following in the diagram

Pt(s)

Cl2(g)

Ag(s)

NaCl(aq)

AgNO3(aq)

Pt 2+, 4+, 6+ Though it states Pt is inert

Cl 2-

Ag 1+

Na 1+

NO3-

Anode definition: the positively charged electrode by which the electrons leave an electrical device.

Electrode definition: a conductor through which electricity enters or leaves an object, substance, or region.

Cations attracted to cathode pick up electrons

Anions attracted to anode release electrodes+

Reduction at Cathode (red cat gain of e)

Oxidation at Anode (ox anode loss of e)

So from the diagram we can see that the charge is being generated through the 2 metal plates.

So the answer is A, the anode material is Pt and the half reaction is 2Cl- = Cl2 + 2e-

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Answer: 0.9375 g

Explanation:

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

$\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$ .....(1)

Molarity of $HCl$ solution = 0.75 M

Volume of $HCl$ solution = 25.0 mL = 0.025 L

Putting values in equation 1, we get:

$\text{Moles of} HCl={0.75}\times{0.025}=0.01875moles$

$CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(s)+CO_2(g)+H_2O(l)$

According to stoichiometry :

2 moles of $HCl$ require = 1 mole of $CaCO_3$

Thus 0.01875 moles of $HCl$ will require= $\frac{1}{2}\times 0.01875=0.009375moles$ of $CaCO_3$

Mass of $CaCO_3=moles\times {\text {Molar mass}}=0.009375moles\times 100g/mol=0.9375g$

Thus 0.9375 g of $CaCO_3$ is required to react with 25.0 ml of 0.75 M HCl

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