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Trava [24]
2 years ago
5

If 25.0 g NO are produced, how many grams of nitrogen gas are used?

Chemistry
1 answer:
ivanzaharov [21]2 years ago
7 0

Based on the assumption that the reaction involves N and O to produce NO, if 25.0 g of NO are produced, the amount of N gas used would be 11.66 grams

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

       N + O ---------> NO

Mole ratio of N to NO is 1:1

Mole of 25.0 g of NO = 25/30.01 = 0.833 moles

Equivalent mole of N = 0.833 moles

Mass of 0.833 moles N = 0.833 x 14 = 11.66 grams

More on stoichiometric calculations can be found here: brainly.com/question/8062886

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Crazy stuff, it will bubble up and flow over
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Which statement about the gas laws is correct?
Elena-2011 [213]

Answer:

Gay-Lussac's law states that pressure and temperature are directly proportional

Explanation:

Gay-Lussac's law states that pressure and temperature are directly proportional. This always occurs if the volume keeps in constant.

n and V are not directly proportional, they are the same.

At Charles Gay Lussac's law

V1 = V2

n1 = n2

T1 < T2

P1 < P2

P1 / T1 = P2 / T2

If the pressure is contant:

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4 0
3 years ago
Ionic bonds form between which two types of elements?​
Alla [95]

Answer:

Ionic bonds usually occur between metal and nonmetal ions. For example, sodium (Na), a metal, and chloride (Cl), a nonmetal, form an ionic bond to make NaCl.

Explanation:

4 0
3 years ago
Read 2 more answers
A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.
BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\  &=185.59\text{ g/mol} \end{aligned}

Dimensional Analysis:

\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0
3 years ago
Which compound has the same empirical and molecular formula ethyne ethene ethane methane?
Margaret [11]
Empirical formula is the simplest ratio of whole numbers of components in a compound 
molecular formula is the actual ratio of components in a compound .
the molecular formula for the compounds given are as follows
ethyne  - C₂H₂
ethene - C₂H₄
ethane - C₂H₆
methane - CH₄
the actual ratios of the elements                 simplified ratio
                     C : H                                           C : H
ethyne            2:2                                             1:1
ethene            2:4                                             1:2
ethane            2:6                                             1:3
methane         1:4                                             1:4
the only compound where the actual ratio is equal to the simplified ratio is methane 
therefore in methane molecular formula CH₄ is the same as empirical formula CH₄
6 0
3 years ago
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