The modern synthesis combined the concepts of ______

The heat of fusion for water is 80. cal/g. How many calories of heat are needed to melt a 35 g ice cube that has a temperature o

HACTEHA [7]

Answer: 2800 calories

Explanation:

Latent heat of fusion is the amount of heat required to convert 1 mole of solid to liquid at atmospheric pressure.

Amount of heat required to fuse 1 gram of water = 80 cal

Mass of ice given = 35 gram

Heat required to fuse 1 g of ice at $0^0C$ = 80 cal

Thus Heat required to fuse 35 g of ice = $\frac{80}{1}\times 35=2800cal$

Thus 2800 calories of energy is required to melt 35 g ice cube

6 0

3 years ago

A 25.0 L tank of nitrogen gas is at 25 oC and 2.05 atm . If the temperature stays at 25 oC and the volume is decreased to 14.5 L

NeTakaya

Answer:

$\boxed {\boxed {\sf P_2 \approx 3.53 \ atm}}$

Explanation:

In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:

$P_1V_1=P_2V_2$

Now we can substitute any known values into the formula.

Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.

$25.0 \ L * 2.05 \ atm = P_2V_2$

The volume is decreased to 14.5 liters, but the pressure is unknown.

$25.0 \ L * 2.05 \ atm = P_2 * 14.5 \ L$

Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .

$\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}=\frac{P_2 *14.5 \ L}{14.5 \ L}$

$\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}= P_2$

The units of liters cancel.

$\frac {25.0 * 2.05 \ atm }{14.5 }=P_2$

$\frac {50.25\ atm }{14.5 }=P_2$

$3.53448276 \ atm = P_2$

The original values of volume and pressure have 3 significant figures, so our answer must have the same.

For the number we found, that is the hundredth place.

3.53448276

The 4 in the thousandth place (in bold above) tells us to leave the 3 in the hundredth place.

$3.53 \ atm \approx P_2$

The new pressure is approximately 3.53 atmospheres.

8 0

2 years ago

a) Whatis the composition in mole fractions of a solution of benzene and toluene that has a vapor pressure of 35 torr at 20 °C?

iogann1982 [59]

Answer:

molar composition for liquid

xb= 0.24

xt=0.76

molar composition for vapor

yb=0.51

yt=0.49

Explanation:

For an ideal solution we can use the Raoult law.

Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.

For toluene and benzene would be:

$P_{B}=x_{B}*P_{B}^{o}$

$P_{T}=x_{T}*P_{T}^{o}$

Where:

$P_{B}$ is partial pressure for benzene in the liquid

$x_{B}$ is benzene molar fraction in the liquid

$P_{B}^{o}$ vapor pressure for pure benzene.

The total pressure in the solution is:

$P= P_{T}+ P_{B}$

And

$1=x_{B}+x_{T}$

Working on the equation for total pressure we have:

$P=x_{B}*P_{B}^{o} + x_{T}*P_{T}^{o}$

Since $x_{T}=1-x_{B}$

$P=x_{B}*P_{B}^{o} + (1-x_{B})*P_{T}^{o}$

We know P and both vapor pressures so we can clear $x_{B}$ from the equation.

$x_{B}=\frac{P- P_{T}^{o}}{ P_{B}^{o} - P_{T}^{o}}$

$x_{B}=\frac{35- 22}{75-22} = 0.24$

So

$x_{T}=1-0.24 = 0.76$

To get the mole fraction for the vapor we know that in the equilibrium:

$P_{B}=y_{B}*P$

$y_{T}=1-y_{B}$

So

$y_{B} =\frac{P_{B}}{P}=\frac{ x_{B}*P_{B}^{o}}{P}$

$y_{B}=\frac{0.24*75}{35}=0.51$

$y_{T}=1-0.51=0.49$

Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.

7 0

3 years ago

A mixture of 5L of H2 and 3L O2 reacts to form H2O (g) at constant T and P .Find the

slamgirl [31]

The volume of H₂O = 5 L

<h3>Further explanation</h3>

Given

5L of H₂ and 3L O₂

Reaction

2H₂ (g) + O₂(g) ⇒2H₂O(g)

Required

The volume of H₂O

Solution

Avogadro's hypothesis:

In the same T,P and V, the gas contains the same number of molecules

So the ratio of gas volume will be equal to the ratio of gas moles

mol H₂ = 5, mol O₂ = 3

Find limiting reactants

From equation, mol ratio H₂ : O₂ = 2 : 1, so :

$\tt H_2\div O_2=\dfrac{mol~H_2}{coefficient}\div \dfrac{mol~O_2}{coefficient}\\\\=\dfrac{5}{2}\div \dfrac{3}{1}=2.5\div 3\rightarrow H_2~limiting~reactant(smaller~ratio)$

Find volume H₂O

mol H₂O based on mol H₂, and from equation mol ratio H₂ : H₂O=2 : 2, so mol H₂O = 5 mol and the volume also 5 L

4 0

2 years ago

During laparoscopic surgery , carbon dioxide gas is used to expand the abdomen to help create a larger working space. If 4.80 L

Studentka2010 [4]

Answer:

5.37 L

Explanation:

To solve this problem we need to use the PV=nRT equation.

First we calculate the amount of CO₂, using the initial given conditions for P, V and T:

P = 785 mmHg ⇒ 785/760 = 1.03 atm

V = 4.80 L

T = 18 °C ⇒ 18 + 273.16 = 291.16 K

1.03 atm * 4.80 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 291.16 K

We solve for n:

n = 0.207 mol

Then we use that value of n for another PV=nRT equation, where T=37 °C (310.16K) and P = 745 mmHg (0.98 atm).

0.98 atm * V = 0.207 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 310.16 K

And we solve for V:

V = 5.37 L

7 0

3 years ago

The modern synthesis combined the concepts of _______ and evolution.