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Verdich [7]
3 years ago
9

Read the passage from "Ode on a Grecian Urn." What men or gods are these? What maidens loth? What mad pursuit? What struggle to

escape? What pipes and timbrels? What wild ecstasy? What is the impact of Keats's repetition of the word what? It explains the overall theme. It makes the poem more formal. It emphasizes the curious tone of the speaker. It expresses the thoughts of the people on the urn.
Chemistry
2 answers:
IRISSAK [1]3 years ago
5 0

Answer:

the correct answer is C

NeX [460]3 years ago
4 0
I believe the correct answer is <span>that it emphasizes the curious tone of the speaker.
The speaker is interested in what is going on on that urn, what is depicted there and why. He wants to have answers to many of his questions because he is curious, and this is why he is constantly repeating the interrogative word what.
</span>
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How many liters of oxygen gas, at standard temperature and pressure, will react with 25.0 grams of magnesium metal? Show all of
love history [14]
Given:
2 Mg + O2 → 2 MgO 
So,
<span>(25.0 g of Mg / 24.3051 g/mole) x (1/2) x (22.4 L/mole) = 11.5 L  of Oxygen will react with 25 grams of magnesium metal.</span>
8 0
3 years ago
Read 2 more answers
A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling
aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}        .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL

Putting values in above equation, we get:

0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g

Hence, the mass of glucose in final solution is 0.420 grams

3 0
3 years ago
Determine the vapor pressure (atm) of rubbing alcohol (isopropanol) at 20.0 °C. The normal boiling point of isopropanol is 82.3
jeyben [28]
Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)

The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)

ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
P₂ = 101.2 kPa
8 0
3 years ago
How many grams of NH3 are needed to provide the same number of molecules as in 0.85 grams of SF6?
solniwko [45]

Above it says the molecular weights are

NH3- 17g/mol and SF6-146 g/mol

Well 1 mole of SF6 is 146.048 grams (i added hte atomic masses of each element). So then the number of moles in 0.85 grams would be 0.00582000438 moles.


<span><span><span>= 1mole / </span><span>146.048g *</span></span> 0.85g</span>


so we would need 0.00582000438 moles of NH3 to have the same number of molecules.

One mole of NH3 is 17.030519999989988 grams (i added each atoms mass). so 0.00582000438 moles of NH3 would be:


<span><span><span>= 17.030519999989988 g / </span><span>mole * </span></span>0.00582000438moles</span>


that equals 0.09911770099 grams.

so 0.09911770099 grams is the answer if you round that you get about 0.1 grams


3 0
3 years ago
Read 2 more answers
1. Oxygen was discovered by Joseph Priestley in 1774 when he heated mercury (II) oxide, HgO, to decompose it to form its constit
Tomtit [17]

Answer:

1. 7.81 moles HgO

2. n = mass/molar mass = (4000 g)/(159.69 g/mol) = 25.05 mol.

Explanation:

How many moles of mercury (II) oxide are needed to produce 125 g of oxygen?

2HgO ==> 2Hg + O2  

125 g O2 x 1 mol O2/32 g x 2 mol HgO / mol O2 = 7.81 moles HgO

------------------------------------------------------------------------------------------------------------

If 4000 g of Fe2O3 is available to react, how many moles of CO are needed?

The no. of moles of CO are needed = 75.15 mol.

Fe₂O₃ + 3CO → 2Fe + 3CO₂,

It is clear that 1 mol of Fe₂O₃ reacts with 3 mol of CO to produce 2 mol of Fe and 3 mol of CO₂.

If 4.00 kg Fe₂O₃ are available to react, how many moles of CO are needed?

We need to calculate the no. of moles of 4.00 kg Fe₂O₃:

n = mass/molar mass = (4000 g)/(159.69 g/mol) = 25.05 mol.

Using cross multiplication:

1 mol of Fe₂O₃ need  → 3 mol of CO to react completely, from stichiometry.

25.05 mol of Fe₂O₃ need  → ??? mol of CO to react completely.

The no. of moles of CO are needed = (3 mol)(25.05 mol)/(1 mol) = 75.15 mol.

6 0
2 years ago
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