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marissa [1.9K]
3 years ago
14

The

Chemistry
1 answer:
julia-pushkina [17]3 years ago
7 0
Octet rule (quizlet helps)
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Arrange the metals used in this experiment (including H2 and Ag) in order of decreasing reactivity (most reactive metal first).
jarptica [38.1K]

Answer:

Na, Ca, Mg, All, Zn, Fe, Sn, Pb, H2, Cu, Ag

Explanation:

This is how metals are arranged in the electrochemical series.

Use the following to remember the order:

King Nathan Can Manage All Zone Freely Since Probably He Can Handle All Ages

Each word represent the order

K, Na, Ca, Mg, All, Zn, Fe, Sn, Pb, H2, Cu, Hg, Ag, Au

6 0
4 years ago
12500 J/ (106 g) (4C) reduce to one
Sloan [31]

The given expression is \frac{12500 J}{(106g)(4^{0}C )}

This expression denotes the specific heat of a substance which indicates a heat energy of 12500 J is involved in raising the temperature of 106 g of the substance by 4^{0}C. Generally, the units of specific heat are \frac{J}{g.^{0}C }

\frac{12500 J}{(106g)(4^{0}C )} = \frac{12500 J}{(106g)(4^{0}C } = 29.5 \frac{J}{g^{0}C }

Therefore,  \frac{12500 J}{(106g)(4^{0}C )} when reduced to one unit will be 29.5 \frac{J}{g^{0}C }

5 0
3 years ago
Read 2 more answers
Why do canned baked beans last longer in a can than in air?
Alinara [238K]
Because when anything is exposed to air, it come in contact with bacteria which begin to break down the proteins in the beans. But when it’s in a can, it can’t get exposed to bacteria which means it won’t break down.

Hope this sheds some light ♥︎

8 0
3 years ago
Read 2 more answers
The combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110 kg of carbon dioxide. What is the limiti
nadezda [96]

Answer:

The limiting reactant is the propane gas, C₃H₈ while the percentage yield is 83.77%

Explanation:

Here we have

Propane gas with molecular formula C₃H₈, molar mass  = 44.1 g/mol combining with O₂ as follows

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Therefore, 1 mole of C₃H₈  combines with 5 moles of O₂ to produce 3 moles CO₂ and 4 moles of H₂O

Mass of propane = 0.1240 kg = 124.0 g

Number of moles of propane = mass of propane/(molar mass of propane)

The number of moles of propane = 124/44.1 = 2.812 moles

The molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.3110 kg = 311.0 g

Therefore, number of moles of CO₂ = mass of CO₂/(molar mass of CO₂)

The number of moles of CO₂ = 311.0 kg/ 44.01 g/mol = 7.067 moles

Therefore, since 1 mole of propane produces 3 moles of CO₂, 2.812 moles of propane will produce 3 × 2.812 moles or 8.44 moles of CO₂

Therefore;

The limiting reactant is the propane gas, C₃H₈, since the oxygen is in excess

Hence

The \ percentage \ yield = \frac{Actual \, yield}{Theoretical \, yield} \times 100 = \frac{7.067}{8.44} \times 100 = 83.77 \%

The percentage yield = 83.77%.

7 0
3 years ago
Please help :,) I’m being timed
vova2212 [387]

Answer:

B

Explanation:

6 0
3 years ago
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