All categories

2 years ago

How do moss leaves and fish differ? How are they the same?

1 answer:

4 0

Moss leaves and fish are different in that, the moss leave is a producer, that is, it produces its own food through photosynthesis while the fish is a consumer, it feeds on foods that are not produced by it.
Both moss and fish are the same in the sense that both have cell as their basic unit of life, that is, they both possess cells.

You might be interested in

El cuerpo humano tiene unos 6 billones de células (6,0x1012) y la población de la Tierra es de unos 8 000 millones de personas.

zloy xaker [14]

Answer:

En toda la población del mundo hay 0.0797 moles de células

Explanation:

1.0 mol of cells = 6.022 * 10∧23 cells

X mol of cells = 6.0 * 10∧12 cells

- X is cleared to find out how many moles of cells are in a human body:

X = 6.0 * 10∧12 cells / 6.022 * 10∧23 cells

X = 9,963 * 10∧-12 moles of cells per person

In the world there are 8 * 10∧9 people, how many moles in total will there be?

8 * 10∧9 people * 9,963 * 10∧-12 moles of cells per person =

0.0797 moles of molecules in the entire population of the earth.

4 0

2 years ago

Silver nitrate and aluminum chloride react with each other by exchanging anions: 3agno3 (aq) + alcl3 (aq) → al(no3)3 (aq) + 3agc

lions [1.4K]

Silver nitrate and aluminum chloride react with each other by exchanging anions: 3agno3 (aq) + alcl3 (aq) → al(no3)3 (aq) + 3agcl (s) what mass in grams of agcl is produced when 4.22 g of agno3 react with 7.73 g of alcl3?

5 0

2 years ago

Read 2 more answers

Diborane, B2H6 a possible rocket propellant, can be made by using lithium hydride (LiH): 6 LiH+ 2 BCl2àB2H6+ 6 LiCl . If you mix

Genrish500 [490]

Answer :

(a) Limiting reactant = $LiH$

(b) The excess reactant = $BCl_3$

(c) The percent of excess reactant is, 50.87 %

(d) The percent yield of $B_2H_6$ or percent conversion of $LiH$ to $B_2H_6$ is, 38.80 %

(e) The mass of $LiCl$ produced is, 1066.42 lb

Explanation : Given,

Mass of $LiH$ = 200 lb = 90718.5 g

conversion used : (1 lb = 453.592 g)

Mass of $BCl_3$ = 1000 lb = 453592 g

Molar mass of $LiH$ = 7.95 g/mole

Molar mass of $BCl_3$ = 117.17 g/mole

Molar mass of $B_2H_6$ = 27.66 g/mole

Molar mass of $LiCl$ = 42.39 g/mole

First we have to calculate the moles of $LiH$ and $BCl_3$ .

$\text{Moles of }LiH=\frac{\text{Mass of }LiH}{\text{Molar mass of }LiH}=\frac{90718.5g}{7.95g/mole}=11411.13moles$

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{453592g}{117.17g/mole}=3871.23moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$6LiH+2BCl_3\rightarrow B_2H_6+6LiCl$

From the balanced reaction we conclude that

As, 6 moles of $LiH$ react with 1 mole of $BCl_3$

So, 11411.13 moles of $LiH$ react with $\frac{11411.13}{6}=1901.855$ moles of $BCl_3$

From this we conclude that, $BCl_3$ is an excess reagent because the given moles are greater than the required moles and $LiH$ is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 3871.23 - 1901.855 = 1969.375 moles

Total excess reactant = 3871.23 moles

Now we have to determine the percent of excess reactant $(BCl_3)$ .

$\% \text{ excess reactant}=\frac{\text{Moles of remaining excess reactant}}{\text{Moles of total excess reagent}}\times 100$

$\% \text{ excess reactant}=\frac{1969.375}{3871.23}\times 100=50.87\%$

The percent of excess reactant is, 50.87 %

Now we have to calculate the moles of $B_2H_6$ .

As, 6 moles of $LiH$ react to give 1 mole of $B_2H_6$

So, 11411.13 moles of $LiH$ react to give $\frac{11411.13}{6}=1901.855$ moles of $B_2H_6$

Now we have to calculate the mass of $B_2H_6$ .

$\text{Mass of }B_2H_6=\text{Moles of }B_2H_6\times \text{Molar mass of }B_2H_6$

$\text{Mass of }B_2H_6=(1901.855mole)\times (27.66g/mole)=52605.3093g$

Now we have to calculate the percent yield of $B_2H_6$ .

$\%\text{ yield of }B_2H_6=\frac{\text{Actual yield of }B_2H_6}{\text{Theoretical yield of }B_2H_6}\times 100=\frac{20411.7g}{52605.3093g}\times 100=38.80\%$

The percent yield of $B_2H_6$ or percent conversion of $LiH$ to $B_2H_6$ is, 38.80 %

Now we have to calculate the moles of $LiCl$ .

As, 6 moles of $LiH$ react to give 6 mole of $LiCl$

So, 11411.13 moles of $LiH$ react to give 11411.13 moles of $LiCl$

Now we have to calculate the mass of $LiCl$ .

$\text{Mass of }LiCl=\text{Moles of }LiCl\times \text{Molar mass of }LiCl$

$\text{Mass of }LiCl=(11411.13mole)\times (42.39g/mole)=483717.8007g=1066.42lb$

The mass of $LiCl$ produced is, 1066.42 lb

8 0

2 years ago

How many grams of hydrogen are produced if 30.0 g of zinc reacts?

alekssr [168]

0.925 grams if using hydrochloric acid in the reaction. 0.462 grams if using sulfuric acid in the reaction. 0.000 grams if using nitric acid in the reaction. Assuming you're using HCl or a similar acid for this reaction, the equation for the reaction is: Zn + 2 HCl ==> ZnCl2 + H2 So each mole of zinc used, produces 1 mole of hydrogen gas, or 2 moles of hydrogen atoms. So we need to look up the atomic weights of both zinc and hydrogen. Atomic weight zinc = 65.38 Atomic weight hydrogen = 1.00794 Moles zinc = 30.0 g / 65.38 g/mol = 0.458855919 mol Since we produce 2 moles of hydrogen atoms per mole of zinc, multiply by 2 and the atomic weight of hydrogen to get the mass of hydrogen produced. So 0.458855919 * 2 * 1.00794 = 0.92499847 grams. Rounding to 3 significant figures gives 0.925 grams. To show the assumption of the acid used, the balanced equation for sulfuric acid would be Zn2 + H2SO4 ==> Zn(SO4)2 + H2 Which means that for every mole of zinc used, 1 mole of hydrogen gas is generated (half that produced via hydrochloric acid). If nitric acid were used, the reaction is 4Zn + 10HNO3 ==> 4Zn(NO3)2 + N2O + 5H2O Which means that NO hydrogen gas is generated. The only justification for assuming hydrochloric acid is used is that it's a fairly common acid that's easy to obtain. But as shown above with 2 alternative acids, the amount of hydrogen gas generated is very dependent upon the exact chemical reaction occurring and asking "How many grams of hydrogen are produced if 30.0 g of zinc reacts?" is a rather silly question unless you specify EXACTLY what the reaction is.

3 0

2 years ago

choose the reaction that illustrates delta H *f for Ca(NO3)2.(A) Ca (s) + N2 (g) + 3O2 ---> Ca(NO3)2 (s)(B) Ca2 (aq) + 2 NO3-

kompoz [17]

Answer: The correct answer is Option A.

Explanation:

Standard enthalpy of formation is the change in enthalpy of one mole of a substance present at the standard state that is 1 atm of pressure and 298 K of temperature. The substance is formed from its pure elements under the same conditions.

We are given a chemical compound having chemical formula $Ca(NO_3)_2$

This compound is formed by the combination of calcium, nitrogen and oxygen elements.

The chemical equation for the formation of $Ca(NO_3)_2$ from the components in their standard states follows:

$Ca+N_2+3O_2\rightarrow Ca(NO_3)_3$

Hence, the correct answer is Option A.

3 0

2 years ago