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snow_lady [41]
3 years ago
5

A certain electrochemical cell has for its cell reaction: zn + hgo → zno + hg which is the half-reaction occurring at the anode?

Chemistry
1 answer:
Leya [2.2K]3 years ago
3 0
It is b because I'm just going around the house with me to the mall to get u at the moment I don't have any plans tonight or naw
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Which step is this ?prophase I, Anaphase I, prophase II, Anaphase II
sweet-ann [11.9K]

Answer:

Prophase 1

Explanation:

4 0
3 years ago
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How many grams of potassium bromide, KBr, are in 100mL of a 0.50 M solution?
VMariaS [17]

Answer:

5.95g

Explanation:

1 dm^{3} = 1000 mL

∴ 100 mL = 100 ÷ 1000 = 0.1 dm^{3}

Volume = <u>0.1 </u>dm^{3}<u />

Concentration = <u>0.5 M</u>

Concentration = \frac{No. of moles}{volume}

0.5 = \frac{x}{0.1}

No. of moles = 0.5 x 0.1 = <u>0.05 moles</u>

No. of moles = \frac{mass}{mass. in. 1. mole}

Mass in 1 mole of KBr = 39 + 80 = 119g (39 is the mass of potassium and 80 is the mass of bromine)

0.05 = \frac{x}{119}

x = 119 × 0.05 = 5.95g

4 0
3 years ago
An object has a mass of 180 kg and a volume of 90 m3. what is its density?
Murljashka [212]
Divide 180kg and 90m3 and your density will be 2kg/m3
8 0
3 years ago
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NaCl + H2SO4 ---&gt; Na2SO4 + HCl<br> Balance the double replacement chemical reaction.
VikaD [51]

Answer:

2NaCl+H2SO4-->Na2SO4+2HCl

Explanation:

There are two Na on the right, so put a 2 in front of NaCl on the left. This makes 3 Cl also, so put a 2 in front of HCl on the right. There are already 2 H on the left, so the equation is balanced.

7 0
4 years ago
For the following reaction, 5.22 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 12.9 grams of
Mamont248 [21]

Answer:

m_{Al_2(SO_4)_3}=17.5gAl_2(SO_4)_3

Explanation:

Hello,

In this case, the undergoing balanced chemical reaction is:

Al_2O_3(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2O(l)

Thus, as 5.22 grams of aluminium oxide reacts, the required yielded amount of aluminium sulfate results:

m_{Al_2(SO_4)_3}=5.22gAl_2O_3*\frac{1molAl_2O_3}{102gAl_2O_3}*\frac{1molAl_2(SO_4)_3}{1molAl_2O_3}*\frac{342gAl_2(SO_4)_3}{1molAl_2(SO_4)_3} \\\\m_{Al_2(SO_4)_3}=17.5gAl_2(SO_4)_3

Moreover, the percent yield is:

Y=\frac{12.9g}{17.5g} *100\%=73.7\%

Best regards.

6 0
4 years ago
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