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3 years ago

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Chemistry

1 answer:

Studentka2010 [4]3 years ago

7 0

Answer:

1. r.a.m = 86×10/100 + 87× 7/100 + 88× 83/100.

860/100 +_607/100 + 7304/100.

8.6+6.07+72.16.

=86.83

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The balanced equation for water is 2 H2 + O2 to 2 H2O. If I have 21.2g of a product , and I started with 5.6 g of H2, how many g

ANTONII [103]

Since 21.2 g H2O was produced, the amount of oxygen that reacted can be obtained using stoichiometry. The balanced equation was given: 2H₂ + O₂ → 2H₂O and the molar masses of the relevant species are also listed below. Thus, the following equation is used to determine the amount of oxygen consumed.

Molar mass of H2O = 18 g/mol

Molar mass of O2 = 32 g/mol

21.2 g H20 x 1 mol H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 18.8444 g O2

We then determine that 18.84 g of O2 reacted to form 21.2 g H2O based on stoichiometry. It is important to note that we do not need to consider the amount of H2 since we can derive the amount of O2 from the product. Additionally, the amount of H2 is in excess in the reaction.

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4 years ago

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

For process 1 :

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

For process 2 :

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

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