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boyakko [2]
4 years ago
15

Smallest to largest: biome,population,organism,ecosystem,community

Chemistry
1 answer:
Ivahew [28]4 years ago
7 0
Organism, population, community, ecosystem, and biome
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How many grams are in 3.4 moles of Lead (IV) Sulfate (Pb(SO4)2)? ) Molar mass: 303.26 g/mol 131.0840​
Murrr4er [49]

Answer:

1031.084g

Explanation:

n=m/M

m=n(M)

m = (3.4mol)(303.26g/mol)

1031.084g

Hope that helps

4 0
2 years ago
Read 2 more answers
Ethyl iodide (C2H5I) decomposes at a certain temperature in the gas phase as follows: C2H5I(g) → C2H4(g) + HI(g) From the follow
lutik1710 [3]

Answer:

Zero order

Explanation:

Looking at the data we can note a linear dependence between concentration and time.  

Time   Conc.

0          2

15         1.82

30        1.64

48        1.42

75        1.10

In the first 15 min it was consumed 2-1.82=0.18. So the rate is r=\frac{\Delta C}{\Delta t} = \frac{0.18}{15}=0.012

From 15 to 30 min (it has passed 15 min) is consumed 1.82-1.64=0.18, so as in the previous calculation the rate is r=0.012.

From 30 to 48  (it has passed 18 min)the rate is r= \frac{0.22}{18}\approx 0.012  

From 48 to 75  (it has passed 27 min) the rate is r= \frac{0.32}{27}\approx 0.012  

So these results suggest that despite of the ever minor concentration of the reactant the rate is ever the same. Hence the reaction rate could be expressed as r= k^{0} = 0.012 mol L^{-1} min^{-1} that is, the reaction is the zero order respect to C2H5I since it is not depending on concentration of C2H5I.  

7 0
3 years ago
KNO
Artyom0805 [142]

Answer:

that so hard man its okay

5 0
3 years ago
What would be the freezing point of a 1.7 mole aqueous ethylene glycol solution? The freezing point depression constant for wate
nata0808 [166]

Answer:

-3.2^oC

Explanation:

In order to answer this question, we need to be familiar with the law of freezing point depression. The law generally states that mixing our solvent with some particular solute would decrease the freezing point of the solvent.

This may be expressed by the following relationship:

\Delta T_f=iK_fb

Here:

\Delta T_f=T_{initial}-T_{final} is the change in the freezing point of the solvent given its initial and final freezing point temperature values;

i is the van 't Hoff factor (i = 1 for non-electrolyte solutes and i depends on the number of moles of ions released per mole of ionic salt);

K_f is the freezing point depression constant for the solvent;

b=\frac{n_{solute}}{m_{solvent}} is molality of the solute, defined as a ratio between the moles of solute and the mass of solvent (in kilograms).

We're assuming that you meant 1.7-molal solution, then:

b=1.7 m

Given ethylene glycol, an organic non-electrolyte solute:

i=1

The freezing point depression constant:

K_f=1.86^oC/m

Initial freezing point of pure water:

T_{initial}=0.00^oC

Rearrange the equation for the final freezing point and substitute the variables:

T_f=T_o-iK_fb=0.00^oC-1\cdot1.86^oC/m\cdot1.7 m=-3.16^oC

8 0
3 years ago
How many unpaired electrons are present in the ground state of a phosphorus atom? a. 3 0.0
inessss [21]

Answer:

3

Explanation:

You need to remember that to measure the number of unparied electrons in an atom you need to undestard its electron configuration, and the electron configuration of phosphorus is 1s2 2s2 2p6 3s2 3p3, just the last state "3p3" have unpaired electron, and because a p state can fits 6 electrons, and here are only 3, that means that those 3 are unpaired.

6 0
4 years ago
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