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2 years ago

Which force controls the size of an atomic nucleus?.

Chemistry

1 answer:

klemol [59]2 years ago

3 0

Answer:

strong nuclear force

Explanation:

1, a force that acts on charged particles

2, a force that holds atomic nuclei together

3, gravity, weak nuclear, electromagnetic, strong nuclear

4, strong nuclear force

5, Gravity and the electromagnetic force have infinite ranges while the nuclear forces have very small ranges.

100% :)

You might be interested in

When the temperature of the gas changes from cold to hot, the amount of pressure is ___________.

Klio2033 [76]

The pressure would increase. When the temperature change form cold to hot, the gas will find ways to escape from containment. Thus, if it cannot escape that pressure will keep on increasing as the temperature rises.

6 0

3 years ago

Pure chlorobenzene is contained in a flask attached to an open-end mercury manometer. When the flask contents are at 58.3°C, the

Elan Coil [88]

Answer:

The slope is 4661.4K

(B), the intercept is 18.156

The vapor pressure of chlorobenzene is 731.4mmHg

The percentage of chlorobenzene originally in the vapor that condenses is = 99.7%

Explanation:

Two sets of conditions (a and b) are observed for L1 and L2 at two different temperatures.

$T_{a}$ = 58.3°C

$L1_{a}$ = 747mmHg

$L2_{a}$ = 52mmHg

$T_{b}$ = 110°C

$L1_{a}$ = 577mmHg

$L2_{b}$ = 222mmHg

We need to convert the temperatures from Celsius to Kelvin (K)

$T_{a}$ = 58.3°C + 273.2

$T_{a}$ = 331.5k

$T_{b}$ = 110°C + 273.2

$T_{b}$ = 383.2k

We then calculate the vapor pressures of the chlorobenzene at each set of conditions by measuring the difference in the mercury levels.

$P^{0}$ = $P_{atm} - (P_1 -P_2)$

The vapor pressure under the first set of conditions is:

$P^{a}$ = 755mmHg - (747mmHg - 52mmHg)

$P^{a}$ = 60mmHg

The vapor pressure under the second set of conditions is:

$P^{b}$ = 755mmHg - (577mmHg = 222mmHg)

$P^{b}$ = 400mmHg

First question say we should find ΔH(slope) and B(intercept) in the Clausius- Clapeyron equation:

Using the Formula :

$In_p^{0} = \frac{- (delta) H}{RT} + B$

where (delta) H = ΔH

The slope of the $In_p^{0} = \frac{T_1T_2 In (P_2/P_1)}{T_1-T_2}$

Calculating the slope; we have:

$\frac{- (delta) H}{R} = \frac{T_1T_2 In (P_2/P_1)}{T_1-T_2}$

$\frac{- (delta) H}{R}$ = $\frac{331.5 * 383.2 * In (400/60)}{(331.5-383.2)}$

$\frac{- (delta) H}{R}$ = -4661.4k

$\frac{ (delta) H}{R}$ = 4661.4k

The intercept can be derived from the Clausius-Clayperon Equation by making B the subject of the formula. To calculate the intercept using the first set of condition above; we have:

B = $In_p_{1} + \frac{ (delta) H}{RT}$

B = $In 60 + \frac{4661.4}{331.5}$

B = 18.156

Thus the Clausius-Clayperon equation for chlorobenzene can be expressed as:

In P = $\frac{-4661.4k}{T} + 18.156$

In question (b), the air saturated with chlorobenzeneis 130°C, converting he temperature of 130°C to absolute units of kelvin(k) we have;

T = 130°C + 273.2

T = 403.2k

Calculating the vapor pressure using Clausius-Clapyeron equation: we have;

$In_p_{0}$ = $In_p_{0} = \frac{-4661.4}{403.2} + 18.156$

$In_p_{0}$ = 6.595

$p_{0}$ = $e^{6.595}$

$p_{0}$ = 731.mmHg

The vapor pressure of chlorobenzene is 731.4mmHg

The diagram of the flowchart with the seperate vapor and liquid stream can be found in the attached document below.

Afterwards, both the inlet and outlet conditions contain saturated liquid.

From the flowchart, the vapor pressure of chlorobenzene at the inlet and outlet temperatures are known:

$P_1(130^{0}C)$ = 731.44mmHg

$P_1(58.3^{0}C)$ = 60mmHg

To calculate the percentage of the chlorobenzene originally in the vapor pressure that condenses; we must first calculate the mole fractions of chlorobenzene for the vapor inlet and outlet using Raoult's Law:

$y_1 = \frac{P^0 (T)}{P}$

At inlet conditions, the mole fraction of chlorobenzene is:

$y_1 = \frac{731.44}{101.3}*\frac{101.3}{760}$

$y_1 = \frac{0.962 mol cholorobenzene}{mol saturated air}$

At outlet conditions , the mole fraction of chlorobenzene is:

$y_2 = \frac{60}{101.3}*\frac{101.3}{760}$

$y_2 = \frac{0.0789 mol cholorobenzene}{mol saturated air}$

Since there is no reaction, the total balance around the condensation is :

$n_1 = n_2 + n_3$

Let assume, that 100 moles of liquid chlorobenzene (CB) is condensed, therefore the equation becomes:

$n_1 = n_2$ + 100mol

The chlorobenzene balance using the mole fractions calculated above is :

$\frac{0.962mol CB}{mol(air)} * n_1 mol(air) = \frac{0.0789molCB}{mol(air)}*n_2mol(air) + 100CB$

substituting equation (1) into equation above; we have:

$\frac{0.962mol CB}{mol(air)} * n_1 mol(air) = \frac{0.0789molCB}{mol(air)}*(n_2-100)mol(air) + 100CB$

$0.962_n_1 mol = 0.0789_n_1mol + 100mol$

we can solve for n1, i.e ;

n1 = 104.3mol total air

Therefore the moles of chlorobenzene that will produce 100 moles of CB liquid is:

$n_C_B = 104.3 (moles) air * \frac{0.962 mol (CB)}{mol air}$

$n_C_B$ = 100.34mol CB

Now, calculating the percentage of chlorobenzene that condenses: we have;

% CB Condensation = $\frac{100mol}{100.34mol}*100%$

% CB Condensation = 99.7%

The percentage of chlorobenzene originally in the vapor that condenses is 99.7%

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4 0

4 years ago

The density of toluene (C7H8) is 0.867 g/mL, and the density of thiophene (C4H4S) is 1.065 g/mL. A solution is made by dissolvin

nadya68 [22]

Answer:

(a) 0.039

(b) 0.384 M

Explanation:

We have the following data:

d(C₇H₈) = 0.867 g/mL

d(C₄H₄S) = 1.065 g/mL

V(C₇H₈) = 250.0 mL

mass(C₄H₄S) = 8.10 g

(a) The mole fraction of C₄H₄S in the solution is the number of moles of C₄H₄S divided into the total moles of the solution:

X(C₄H₄S) = moles C₄H₄S/ total moles

To calculate the moles, we need the molecular weight (MW) of each compound:

MW(C₄H₄S) = (4 x 12 g/mol) + (4 x 1 g/mol) + 32 g/mol = 84 g/mol

MW(C₇H₈) = (7 x 12 g/mol) + (8 x 1 g/mol) = 92 g/mol

Thus, we calculate the moles of C₄H₄S by dividing the mass into the MW(C₄H₄S):

moles C₄H₄S = mass(C₄H₄S)/MW(C₄H₄S)= 8.10 g/(84 g/mol) = 0.096 moles

Then, we have to calculate the moles of C₇H₈. First, we need the mass, obtained from the product of the density by the volume:

mass(C₇H₈)= d(C₇H₈) x V(C₇H₈) = 0.867 g/mL x 250.0 mL = 216.75 g

Thus, we divide the mass of C₇H₈ into the MW to calculate the moles of C₇H₈:

moles C₇H₈ = mass(C₇H₈)/MW(C₇H₈) = 216.75 g/(92 g/mol) = 2.35 moles

The total moles is obtained from the addition of the moles of the solute (C₄H₄S) and the solvent (C₇H₈):

total moles = moles C₄H₄S + moles C₇H₈ = 0.096 moles + 2.35 moles = 2.45 moles

Finally, we calculate the mole fraction of C₄H₄S:

X(C₄H₄S) = moles C₄H₄S/ total moles = 0.096 moles/2.45 moles = 0.039

(b) The molarity of C₄H₄S is calculated as follows:

M(C₄H₄S) = moles C₄H₄S/1 liter solution

Assuming that the total volume of the solution is the volume of solvent (C₇H₈), we calculate the molarity of C₄H₄S by dividing the moles into the volume of solvent in liters:

V(C₇H₈) = 250.0 mL = 0.250 L

M(C₄H₄S) = 0.096 moles/(0.250 L) = 0.384 mol/M = 0.384 M

(c) Assuming that the volumes of solute and solvent are additive, we can add the volumes of C₄H₄S and C₇H₈. First, we need the volume of C₄H₄S, which can be calculated from the mass and density:

V(C₄H₄S) = mass(C₄H₄S)/ d(C₄H₄S) = 8.10 g/(1.065 g/mL) = 7.606 mL = 0.0076 L

Now, we add the volumes:

total volume = V(C₇H₈) + V(C₄H₄S) = 0.250 L + 0.0076 L = 0.2576 L

Finally, we recalculate the molarity of C₄H₄S:

M(C₄H₄S)= moles C₄H₄S/ total volume = 0.096 moles/0.2576 L = 0.373 M

7 0

3 years ago

Somebody help with A B and C please will give brain thanks

Norma-Jean [14]

Answer:

we cant see a b or c

Explanation:

7 0

3 years ago

What is the mass of 16800 mL of oxygen gas at STP

Pepsi [2]

Due to having STP conditions and given volume you can use this conversion: 1 mol=22.4L at STP. Since you have mL, convert to Liters first. 16800 mL = 16.8 L. You will need molar mass of Oxygen(O2)

16.8 L O2 x(1 mol/22.4 L) x( 31.998 g O2/1 mol) = 24.0 g O2

Alternatively, maybe you did not know 1 mol =22.4 L conversion, you can use PV=nRT will work, with P = 1 atm, R =0.08206 L(atm)/mol(k), and T=273 K, V = 16.8 L. Solve for n(moles), then multiply by molar mass of O2

6 0

3 years ago