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joja [24]
3 years ago
6

Which is not an example of a state change? wood burning in a fire. water turning to steam. gold melting. carbon dioxide freezing

into dry ice?
Chemistry
1 answer:
eimsori [14]3 years ago
5 0
The only example which is not a state change is wood burning into a fire, which is a <em>chemical</em> change. The remainder of the choices are all changes of state. e.g. changing between solid, liquid, and gas phases, with no chemical reaction.
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Using the correct answer from Part B, calculate the volume of a rectangular prism with a length of 5.6 cm, a width of 2.1 cm, an
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5.6 x 2.1 x 6.6 = 77.616 cm3
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Three beakers contain clear, colorless liquids. One beaker contains pure water, another contains salt water, and another contain
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Pure water may be identified by its density: it will have the lowwest density of the three solutions, given that the solutes (salt and sugar) increase the density (this is the easiests and quickest way).

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6 0
3 years ago
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What is the volume of 0.80 grams of o2 gas at stp? (5 points) group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l
Vladimir [108]

Answer:

0.56L

Explanation:

This question requires the Ideal Gas Law:  PV=nRT where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.

Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:R=0.0821\frac{L\cdot atm}{mol\cdot K}

Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means T=273.15K and P=1atm

Lastly, we must calculate the number of moles of O_2(g) there are.  Given 0.80g of O_2(g), we will need to convert with the molar mass of O_2(g).  Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2:  32g\text{ }O_2=1mol\text{ }O_2

Thus, \frac{0.80g \text{ }O_2}{1} \frac{1mol\text{ }O_2}{32g\text{ }O_2}=0.25mol\text{ }O_2=n

Isolating V in the Ideal Gas Law:

PV=nRT

V=\frac{nRT}{P}

...substituting the known values, and simplifying...

V=\frac{(0.025 mol \text{ }O_2)(0.0821\frac{L\cdot atm}{mol \cdot K} )(273.15K)}{(1atm)}

V=0.56L \text{ } O_2

So, 0.80g of O_2(g) would occupy 0.56L at STP.

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2 years ago
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A. Covalent bonds is the answer
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The correct answer would have to be true
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