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3 years ago

This is the balances equations. C3H8 + 5O2 → 3CO2 + 4H2O How many moles of oxygen are required to produce 37.15 g CO2

1 answer:

4 0

Molar mass O2 = 31.99 g/mol

Molar mass CO2 = 44.01 g/mol

Moles ratio:

C3H8 + 5 O2 = 3 CO2 + 4 H2O

5 x 44.01 g O2 ---------------- 3 x 44.01 g CO2
( mass of O2) ------------------ 37.15 g CO2

mass of O2 = 37.15 x 5 x 44.01/ 3 x 44.01

mass of O2 = 8174.8575 / 132.03

mass of O2 = 61.916 g

Therefore:

1 mole O2 ----------------- 31.99 g
moles O2 -------------------- 61.916

moles O2 = 61.916 x 1 / 31.99

moles = 61.916 / 31.99 => 1.935 moles of O2

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Maria is baking cookies. She mixes water, flour, sugar, baking powder, butter and chocolate chips and then measures the mass of

Mrrafil [7]

Answer:

The law of conservation of mass was obeyed since some of the mass evaporated while the cookies were baking.

Explanation:

Maria mixes water, flour, sugar, baking powder, butter and chocolate chips and then measures the mass of her ingredients.

Initial mass of ingredients = 1053 g

Final mass = 845 g

All these information are not of any great help.

Law of conservation of mass states that:

For any system closed to all transfers of matter and energy, the mass of the system can not change with respect to time, as system's mass cannot change, any mass can neither be added nor be removed.

In this process of baking, considering the cookies as our system, they never formed a closed system. Hence were free for evaporation.

During the process of baking, water from dough must have evaporated and mass must have reduced.

Hence, the law of conservation of mass was not obeyed.

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3 years ago

Read 2 more answers

Ymorist [56]

Answer:

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Explanation:

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3 years ago

When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.

Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

$Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3$

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

$0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2$

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

$0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2$

And the resulting percent yield:

$Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%$

Regards!

4 0

3 years ago

Which ion does an acid contain most off

Elena-2011 [213]

Answer:

Hydrogen +

Explanation:

An acid is a chemical which "wants" to donate some protons, or hydrogen + ions. Since a hydrogen atom is just a proton and an electron, the ion lacking an electron is simply a proton. Hope this helps!

6 0

4 years ago

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A galvanic cell at a temperature of is powered by the following redox reaction: 2Cr3 + 3Ca Suppose the cell is prepared with in

stira [4]

Answer:

2.13 V

Explanation:

The balanced equation of the reaction his;

2Cr^3+(aq) + 3Ca(s) -----> 2Cr(s) + 3Ca^2+(aq)

Since this is a galvanic cell then E°cell must be positive. It implies that calcium will be the anode and chromium will be the cathode since calcium is ahead of chromium in the electrochemical series.

E°anode= -2.87 V

E°cathode= -0.74 V

E°cell= E°cathode -E°anode

E°cell= -0.74 -(-2.87)

E°cell = 2.13 V

8 0

3 years ago