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

How many grams of magnesium are needed to completely react

Verizon [17]

97.22 grams of magnesium is needed.

Explanation:

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 calculate the inital number of moles of each reagent, using the given volumes and concentrations:

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 calculate how many Pb(NO₃)₂ moles reacted with 16.0 mmoles of NaCl, using the stoichiometric coefficients of the reaction:

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

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

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

Finally we divide the number of moles by the final volume to calculate the concentration:

5.30 mmol / (52.1 mL + 38.5 mL) = 0.0585 M

6 0

3 years ago