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2 years ago

Compare the way that frogs breathe when they are tadpoles with the way frogs breathe when they are adults.

2 answers:

7 0

Answer:Frogs breathe through gills underwater to get oxygen and when frogs are adults they breathe through their lungs to get oxygen.

Explanation:

andreev551 [17]2 years ago

6 0

When frogs are tadpoles they, they breathe through their gills, as frogs they have mucus glands in their skin which helps them absorb oxygen from the air, or they can also breathe through their lungs

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Jerome is learning how the model of the atom has changed over time as new evidence was gathered. He has images of four models of

stepladder [879]

Answer:

the answer is A. Y, X, Z, W

Explanation:

I took the test

8 0

3 years ago

Read 2 more answers

Propane has a normal boiling point of -42.04 °C and a heat of vaporization of 24.54 kJ/mole. What is the vapor pressure of propa

Mademuasel [1]

Answer : The vapor pressure of propane at $25.0^oC$ is 17.73 atm.

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of propane at $25.0^oC$ = ?

$P_2$ = vapor pressure of propane at normal boiling point = 1 atm

$T_1$ = temperature of propane = $25.0^oC=273+25.0=298.0K$

$T_2$ = normal boiling point of propane = $-42.04^oC=230.96K$

$\Delta H_{vap}$ = heat of vaporization = 24.54 kJ/mole = 24540 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{1atm}{P_1})=\frac{24540J/mole}{8.314J/K.mole}\times (\frac{1}{298.0K}-\frac{1}{230.96K})$

$P_1=17.73atm$

Hence, the vapor pressure of propane at $25.0^oC$ is 17.73 atm.

3 0

3 years ago

For the reaction of ammonia (NH3) with oxygen (O2) to produce water and nitric oxide (NO), how many moles of water are produced

asambeis [7]

Answer:

3.3 moles of H₂O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4NH₃ + 5O₂ —> 6H₂O + 4NO

From the balanced equation above,

4 moles of NH₃ reacted to produce 6 moles of H₂O.

Finally, we shall determine the number of mole of H₂O produced by the reaction of 2.2 moles of NH₃. This can be obtained as follow :

From the balanced equation above,

4 moles of NH₃ reacted to produce 6 moles of H₂O.

Therefore, 2.2 moles of NH₃ will react to produce = (2.2 × 6)/4 = 3.3 moles of H₂O.

Thus, 3.3 moles of H₂O were obtained from the reaction.

7 0

2 years ago

What is the name of the compound Ca(NO3)2 ?

natulia [17]

Answer: B) calcium nitrate

5 0

3 years ago

Mg(OH)2 + 2 HBr à MgBr2 + 2 H2O

AnnyKZ [126]

Explanation:

The balanced equation of the reaction is given as;

Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)

1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?

From the reaction;

2 mol of HBr produces 1 mol of MgBr2

Converting to masses using;

Mass = Number of moles * Molar mass

Molar mass of HBr = 80.91 g/mol

Molar mass of MgBr2 = 184.113 g/mol

This means;

(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2

18.3g would produce x

161.82 = 184.113

18.3 = x

x = (184.113 * 18.3 ) / 161.82 = 20.8 g

2. How many moles of H2O will be produced from 18.3 grams of HBr?

Converting the mass to mol;

Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol

From the reaction;

2 mol of HBr produces 2 mol of H2O

0.226 mol would produce x

2 =2

0.226 = x

x = 0.226 * 2 / 2 = 0.226 mol

3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?

From the reaction;

2 mol of HBr reacts with 1 mol of Mg(OH)2

18.3g of HBr = 0.226 mol

2 = 1

0.226 = x

x = 0.226 * 1 /2

x = 0.113 mol

5 0

3 years ago