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tensa zangetsu [6.8K]
3 years ago
15

for the reactants represented by the equation 2H2 + O2 > 2H2O, how many grams of water are produced from 376.97 mol of hydrog

en?
Chemistry
2 answers:
Hunter-Best [27]3 years ago
8 0

Answer:

6792.1 g

Explanation:

Dimensional analysis:

376.92 mol H2 * (2 mol h2o / 2 mol h2) * (18.02g h2o / 1 mol h2o)

6792.1 g

kirza4 [7]3 years ago
6 0

Answer:

Amount of water produced from 376.97 mole hydrogen is 6785.46 g

Explanation:

Given number of moles of hydrogen = 376.97 mole

The balanced reaction is shown below

2\textrm{H}_{2}\left ( g \right )+\textrm{O}_{2}\left ( g \right )\rightarrow 2\textrm{H}_{2}\textrm{O}\left ( l \right )

2 mole of water is produced from 2 mole of hydrogen gas according to the balanced reaction.

Number of moles of water produced by 376.97 mole hydrogen = 376.97 mole

Molar mass of water = 1 8 g/mol

\textrm{Amount of water produced} = \left ( 376.97 \textrm{ mole}\times 18\textrm{ g/mol} \right ) = 6785.46 \textrm{ g}

Amount of water produced = 6785.46 g

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8 0
3 years ago
In a titration, the point at which one drop of base turns the acid indicator a pink color that lasts for 30 seconds is called th
xxTIMURxx [149]
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Kazeer [188]
The answer is c this is the answer
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An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i
lutik1710 [3]

Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

Best regards.

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