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3 years ago

How are cone bearing trees and static electricity connected

Chemistry

1 answer:

Pavlova-9 [17]3 years ago

5 0

Answer: Static electricity will help in seed dispersal in cone bearing seeds.

Explanation:

Static electricity is the stationary electric charge produced by the friction. The static electricity can be produced by the turbulence or air currents in the storm clouds. This is responsible for the cause of lightening phenomena.

The cone bearing seeds exhibit cones that are hard and woody in nature. The static electricity due to the effect of lightening generate electrical shock which breaks open the cone and disperse the seeds. Thus the static electricity is important for seed dispersal in cone bearing trees.

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What else supports the idea that an exothermic reaction has occurred?

kodGreya [7K]

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3 years ago

The density of whole milk 1.04 g per mL. What is the volume (in quarts) of 18.5 pounds of whole milk?

nadya68 [22]

The volume : 8,526 quarts

<h3>Further explanation</h3>

Given

The density of whole milk = 1.04 g/ml

mass = 18.5 pounds

Required

The volume

Solution

Conversion of mass

1 pound = 453,592 g

18.5 pounds = 8391,45 g

Density formula:

$\large{\boxed {\bold {\rho~=~ \frac {m} {V}}}}$ .

Input the value :

V = m : ρ

V = 8391,45 g : 1.04 g/ml

V = 8068.7 ml

1 ml = 0,00105669 quarts

8068.7 ml =8,526 quarts

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2 years ago

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Elanso [62]

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3 years ago

Read 2 more answers

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$