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murzikaleks [220]
3 years ago
8

2NO(g) + O2 (g) --> 2NO2

Chemistry
1 answer:
Tatiana [17]3 years ago
5 0
Take 5 * 2/1 = 10 mol NO needed
10 mol * 30 g/mol = 300g

Answer C
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Naturally occuring element X exists in three isotopic forms: X-28 (27.730 amu, 60.58% abundance), X-29 (28.841 amu, 18.35% abund
bazaltina [42]

The Average atomic weight of X is 28.7amu

Isotopes are atoms with the same number of protons but differing numbers of neutrons.

Different isotopes have various atomic masses.

The proportion of atoms with a particular atomic mass that can be found in a naturally occurring sample of an element is known as the relative abundance of an isotope.

An element's average atomic mass is computed as a weighted average by multiplying the relative abundances of its isotopes by their respective atomic masses, then adding the resulting products.

Using mass spectrometry, it is possible to determine the relative abundance of each isotope.

The atomic weight of the element will be a weighted average of the isotopes based on the relative abundance:

(27.730 x 0.6058) + (28.841 x 0.1835) + (31.321 x 0.2107) = 16.7988 + 5.2923+ 6.599 = 28.690 = 28.7 amu.

Average atomic weight of X is 28.7amu

Learn more about Average Atomic Weight here

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3 0
1 year ago
Can someone please help me with these questions. image attached
Radda [10]
Question 9. The first one is the smallest. Anything with a negative exponent is going to be less than 1, the .00000241. The exponent tells you the number of zeroes to the right of the decimal point. Farther to right gets smaller and smaller.

Question 10. The last one is true. If the last digit is smaller than 5, drop the digit, and do not change. (If it is a 5 or larger, the digit before it would round up)
7 0
3 years ago
Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2H2S(g) 3O2(g)2H2O(l) 2SO2
Marizza181 [45]

Answer:

\Delta _rH=-1124.14kJ/mol

Explanation:

Hello!

In this case, since the standard enthalpy change for a chemical reaction is stood for the enthalpy of reaction, for the given reaction:

2H_2S(g) +3O_2(g)\rightarrow 2H_2O(l) +2SO_2(g)

We set up the enthalpy of reaction considering the enthalpy of formation of each species in the reaction at the specified phase and the stoichiometric coefficient:

\Delta _rH=2\Delta _fH_{H_2O,liq}+2\Delta _fH_{SO_2,gas}-2\Delta _fH_{H_2S,gas}-3\Delta _fH_{O_2,gas}

In such a way, by using the NIST database, we find that:

\Delta _fH_{H_2O, liq}=-285.83kJ/mol\\\\\Delta _fH_{SO_2, gas}=-296.84kJ/mol\\\\\Delta _fH_{O_2,gas}=0kJ/mol\\\\\Delta _fH_{H_2S,gas}=-20.50kJ/mol

Thus, we plug in the enthalpies of formation to obtain:

\Delta _rH=2(-285.73kJ/mol)+2(-296.84kJ/mol)-2(-20.50kJ/mol)-3(0kJ/mol)\\\\\Delta _rH=-1124.14kJ/mol

Best regards!

8 0
2 years ago
According to the kinetic molecular theory, the particles of an ideal gas
Vera_Pavlovna [14]
A have no potential energy
7 0
3 years ago
How many moles are in 525 g of ammonia, NH3?<br> 8940.75 M<br> 0.03 M<br> 30.83 M
victus00 [196]

Answer:

30.83 M

Explanation:

17.03052 re in one mole. So, if you multiply it by 30.83, you will get 535 g of ammonia.

In fact, the detailed answer is 30.827009392549122.

3 0
2 years ago
Read 2 more answers
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