2H2O → 2H2 + O2

Which best describes the relationship between Haiti’s political instability and the natural disasters it has faced?

Paul [167]

Explanation:

FOR ALL MY POINTS write an argumentative essay on why dog parks are "helpful or harmful"

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8 0

2 years ago

What can help overcome a positive enthalpy of solution and allow a solid

IgorC [24]

Answer: D. An increase in entropy

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an disordered arrangement to an ordered arrangement, the entropy is said to increase and vice versa.

For a reaction to be spontaneous, the enthalpy of the solution must decrease and the entropy must increase.

To overcome a positive enthalpy of solution and allow a solid solute to dissolve in water, an increase in entropy would make the reaction spontaneous as the system would move to a more disordered state.

4 0

2 years ago

One tank of gold fish is fed the normal amount of food once a day. A second tank is fed twice a day. A third tank is fed four ti

Arturiano [62]

The question is incomplete, the complete question is;

One tank of goldfish is feed the normal amount which is once a day, a second tank is fed twice a day, and a third tank is fed four times a day during a 6 week study. The fishes' body fat is recorded daily.

Independent Variable-

Dependent Variable-

Constants

Control Group-

Answer:

A) The amount of food the gold fish receives

B) Body fat of the gold fish

C) -Type of fish used in the study (gold fish)

Time period within which the fishes were fed (Six week period)

Shape and size of tank

D) group of gold fish fed the normal amount

Explanation:

The purpose of the study is to determined the impact of amount of feed on the body fat of gold fish. Hence, the amount of feed is the independent variable while the body fat of the feed is the dependent variable.

The control group receives the normal amount of feed (once a day). The fishes are all gold fish, fed within a six week period. All the tanks were of the same shape and size. These are the constants in the study.

4 0

2 years ago

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

3 years ago

100 POINTS PLS HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

dimulka [17.4K]

Answer:

i need this im struggling with my work

Explanation:

3 0

2 years ago

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