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Dafna1 [17]
3 years ago
9

When 542 mL of O2 gas at 30°C and 1.04 atm

Chemistry
1 answer:
Anastasy [175]3 years ago
5 0
I hope this help you

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If 20.00 mol of hydrogen gas completely reacted with 20.00 mol of chlorine gas
MrMuchimi

Answer:

the last one

Explanation:

8 0
3 years ago
Make a t chart comparing physical and chemical changes
madam [21]
FOLLOWING!!!!!!!!!!!!!!!!!!
8 0
3 years ago
Zn-64 = 48.63%
Ann [662]

Answer:

A = 65.46 u

Explanation:

Given that,

The composition of zinc is as follows :

Zn-64 = 48.63%

Zn-66 = 27.90%

Zn-67 = 4.10%

Zn-68 = 18.75%

Zn-70 = .62%

We need to find the  average atomic mass of the given element. It can be solved as follows :

A=\dfrac{48.63\times 64+27.90\times 66+4.1\times 67+18.75\times 68+0.62\times 70}{100}\\A=65.46\ u

So, the average atomic mass of zinc is 65.46 u.

5 0
3 years ago
Isotopes of the same element differ only in the number of electrons they contain. isotopes of the same element don't usually hav
alexdok [17]
Isotopes of the same element differ only in the number of neutrons an atom has. Its still the same element because the number of protons define an element. Just their neutrons and mass number is different.
4 0
3 years ago
A student dissolves 15.0 g of ammonium chloride(NH4Cl) in 250. 0 g of water in a well-insulated open cup. She then observes the
iren2701 [21]

Answer:  

1) Endothermic.  

2) Q_{rxn}=4435.04J  

3) \Delta _rH=15.8kJ/mol

Explanation:  

Hello there!  

1) In this case, for these calorimetry problems, we can realize that since the temperature decreases the reaction is endothermic because it is absorbing heat from the solution, that is why the temperature goes from 22.00 °C to 16.0°C.  

2) Now, for the total heat released by the reaction, we first need to assume that all of it is released by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

Q_{rxn}=-(15.0g+250.0g)*4.184\frac{J}{g\°C}(16.0-20.0)\°C\\\\ Q_{rxn}=4435.04J    

3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case NH4Cl, we proceed as follows:

\Delta _rH=\frac{ Q_{rxn}}{n}\\\\\Delta _rH= \frac{ 4435.04J}{15.0g*\frac{1mol}{53.49g} } *\frac{1kJ}{1000J} \\\\\Delta _rH=15.8kJ/mol

Best regards!  

Best regards!

4 0
3 years ago
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