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Lubov Fominskaja [6]
3 years ago
5

A) Which of the following reagents would oxidize Cr to Cr2 , but not Fe to Fe3 ?

Chemistry
1 answer:
sdas [7]3 years ago
3 0
<span>The answer is Co2+  This is the </span><span>reagent that would oxidize Cr to Cr2 , but not Fe to Fe3. A</span> metal that lies between Cr and Sn is Cobalt. In order for it to oxidize Cr, it has to be Co+2 (Cobalt cation).
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The feed to a batch process contains equimolar quantities of nitrogen and methane. write an expression for the kilograms of nitr
timama [110]
1) number of moles of N2 = n/2

2) Number of moles of CH4 = n/2

3) Total number of moles of the mixture = n/2 + n/2 = n

4) Kg of N2

mass in grams = number of moles * molar mass

molar mass of N2 = 2 * 14.0 g/mol = 28 g/mol

=> mass of N2 in grams = (n/2) * 28 = 14n

mass of N2 in Kg = mass of N2 in grams * [1 kg / 1000g] = 14n/1000 kg = 0.014n kg

Answer: mass of N2 in kg = 0.014n kg
3 0
3 years ago
How are electric circuits like water in pipes list two ways
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They flow through the inslation of the house .

they both help things travel
8 0
3 years ago
if you react 100g of ammonium chloride with excess calcium oxide, what Is the theoretical yeild (in grams) of ammonia? when you
Maurinko [17]

Answer:

31.78 grams

25.55%

Explanation:

The balanced reaction for ammonium chloride with calcium oxide will be:

2NH4Cl + Ca(OH)2 ---> CaCl2 + 2NH3 + 2H2O

The molecular weight for ammonium chloride(NH4Cl ) is 53.49g/mol, while the molecular weight for ammonium(NH3) is 17g/mol. The number of theoretical yield of ammonia from 100g of ammonium chloride will be:

100g / (53.49g/mol) * 2/2  * 17g/mol= 31.78 grams

If the actual yield is 8.12g, the percent yield will be: 8.12g/31.78g * 100% =25.55%

4 0
3 years ago
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
Ronch [10]

Answer:

M=0.0637M

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

Finally, the resulting molarity in 30.8 mL (0.0308 L):

M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M

Regards.

3 0
3 years ago
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Answer:

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