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Mrrafil [7]
3 years ago
14

I’ll give people points if they answer this plz(math)

Chemistry
1 answer:
mezya [45]3 years ago
8 0
I’m pretty sure the first option is the right answer
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How many grams of magnesium are needed to completely react
Verizon [17]

97.22 grams of magnesium is needed.

<u>Explanation:</u>

The molar ratio of Mg to O2  to make MgO is 2:1

Here, we have 2 moles of O2, therefore we need 4 moles of Mg.

1 mol of Mg = 24.305 g

Therefore

4 moles of Mg = 4 \times 24.305 g

                        = 97.22 g

Hence we need 97.22grams of magnesium are needed to completely reactwith 2.00 mol of O2 in the synthesis reaction that producesmagnesium oxide.

5 0
3 years ago
Both cholesterol and oleic acid are lipids. What property do they have in common?
OLga [1]
Cholesterol and oleic acid are non-polar (hydrophobic) compounds, soluble in organic solvents. Most membrane lipids are amphipathic, having a non-polar end and a polar end. Oleic acid and other fatty acids are hydrophobic (not soluble in water).

Please mark as brainliest if it helps
8 0
3 years ago
In the following reaction 8 grams of ethane are burned and 11 grams of CO2 are collected. What is the percent yield? 2C2H6+7O2=
Alika [10]
47% yield.  
First, let's determine how many moles of ethane was used and how many moles of CO2 produced. Start with the respective atomic weights. 
Atomic weight carbon = 12.0107 
Atomic weight hydrogen = 1.00794 
Atomic weight oxygen = 15.999  
Molar mass C2H6 = 2 * 12.0107 + 6 * 1.00794 = 30.06904 g/mol 
Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol  
Moles C2H6 = 8 g / 30.06904 g/mol = 0.266054387 mol 
Moles CO2 = 11 g / 44.0087 g/mol = 0.249950578 mol  
Looking at the balanced equation, for every 2 moles of C2H6 consumed, 4 moles of CO2 should be produced. So at 100% yield, we should have 0.266054387 / 2 * 4 = 0.532108774 moles of CO2. But we only have 0.249950578 moles, or 0.249950578 / 0.532108774 = 0.46973587 =
46.973587% of what was expected.  
Rounding to 2 significant figures gives 47% yield.
8 0
4 years ago
1.72 moles of NOCI were placed in a 2.50 L reaction chamber
elena-14-01-66 [18.8K]

The equilibrium constant, Kc=0.026

<h3>Further explanation</h3>

Given

1.72 moles of NOCI

1.16 moles of NOCI  remained

2.50 L reaction chamber

Reaction

2NOCI(g) = 2NO(g) + Cl2(g).

Required

the equilibrium constant, Kc

Solution

ICE method

   2NOCI(g) = 2NO(g) + Cl2(g).

I    1.72

C   0.56           0.56         0.28

E   1.16              0.56         0.28

Molarity at equilibrium :

NOCl :

\tt \dfrac{1.16}{2.5}=0.464

NO :

\tt \dfrac{0.56}{2.5}=0.224

Cl2 :

\tt \dfrac{0.28}{2.5}=0.112

\tt Kc=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\\\\Kc=\dfrac{0.224^2\times 0.112}{0.464^2}=0.026

4 0
3 years ago
52.1 mL of aqueous 0.255 M Pb(NO3)2 is mixed with 38.5 mL of 0.415 M NaCl. The equation for the precipitate reaction is: Pb(NO3)
Elan Coil [88]

Answer:

0.0585 M

Explanation:

  • Pb(NO₃)₂ (aq) + 2NaCl (aq) → PbCl₂ (s) + 2NaNO₃ (aq)

First we <u>calculate the inital number of moles of each reagent</u>, using the <em>given volumes and concentrations</em>:

  • 0.255 M Pb(NO₃)₂ * 52.1 mL = 13.3 mmol Pb(NO₃)₂
  • 0.415 M NaCl * 38.5 mL = 16.0 mmol NaCl

Then we <u>calculate how many Pb(NO₃)₂ moles reacted with 16.0 mmoles of NaCl</u>, using the <em>stoichiometric coefficients of the reaction</em>:

  • 16.0 mmol NaCl * \frac{1mmolPb(NO_3)_2}{2mmolNaCl} = 8.00 mmol Pb(NO₃)₂

Now we <u>calculate the remaining number of Pb(NO₃)₂ moles after the reaction</u>:

  • 13.3 mmol - 8.00 mmol = 5.30 mmol Pb(NO₃)₂

Finally we <em>divide the number of moles by the final volume</em> to <u>calculate the concentration</u>:

  • 5.30 mmol / (52.1 mL + 38.5 mL) = 0.0585 M
6 0
3 years ago
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