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Phantasy [73]
3 years ago
13

A student gains 1.0 lb in two weeks. How many grams did he gain?

Chemistry
2 answers:
Verdich [7]3 years ago
7 0
1lb= 453.592g but then to change to 2 significant figures you go down two of the numbers so here it will be 4 and 5 that stands for 450g. So he gained 450g. I hope that helped
OlgaM077 [116]3 years ago
3 0
1 lb=453.592 grams
So he gained 453.592 pounds
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Yo answer my question
igomit [66]

Answer:

C maybe

Explanation:

3 0
3 years ago
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Why are hydrogen, lithium, and sodium classified as reactive elements?
Sergeu [11.5K]
<span>The alkali metals and hydrogen are reactive because they have only one electron to give in order to complete their valence shell. It is easier to give that one electron so when given the opportunity they will. This means they will react with anything polar or willing to take an electron.</span>
4 0
3 years ago
The enthalpy of combustion for octane (C8H18(l)), a key component of gasoline, is -5,074 kJ/mol. This value is the Delta. Hrxn f
Inessa [10]

Combustion can be defined as the reaction of a compound with oxygen. The enthalpy of combustion of octane is \Delta H_{\rm rxn} for \rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O.

<h3>What is the enthalpy of reaction?</h3>

The enthalpy of reaction is the amount of heat energy absorbed or lost by the molecules in the chemical reaction.

The enthalpy of combustion is the amount of heat energy released by the compound in the reaction with oxygen.

The reaction in which heat is liberated with the reaction of a compound with oxygen has an enthalpy of combustion, equivalent to the enthalpy of reaction.

The combustion of octane can be given as:

\rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O

Thus, the reaction has combustion energy equivalent to the enthalpy of the reaction is \rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O. Thus, option B is correct.

Learn more about enthalpy of reaction, here:

brainly.com/question/1657608

6 0
2 years ago
Where is most of the mass found in an atom
IgorLugansk [536]
massive livand that sarah or someone is how u do it
7 0
3 years ago
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Consider a 1260-kg automobile clocked by law-enforcement radar at a speed of 85.5 km/h. If the position of the car is known to w
-BARSIC- [3]
The expected speed is v = 85.5 km/h
 v = 85.5 km/h = (85.5 km/h)*(0.2778 (m/s)/(km/h)) = 23.75 m/s

If there is an uncertainty of 2 meters in measuring the position, then within a 1-second time interval:
The lower measurement for the speed is v₁ = 21.75 m/s,
The upper measurement for the speed is v₂ = 25.75 m/s.
The range of variation is
Δv = v₂ - v₁ = 4 m/s

The uncertainty in measuring the speed is
Δv/v = 4/23.75 = 0.1684 = 16.84%

Answer: 16.8%
3 0
3 years ago
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