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How can you tell a chemical change has happened?

Paladinen [302]

Answer:

How can I tell if a chemical reaction is occurring? A chemical reaction is usually accompanied by easily observed physical effects, such as the emission of heat and light, the formation of a precipitate, the evolution of gas, or a color change.

Explanation:

4 0

3 years ago

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A sample of PCl5(g) was placed in an otherwise empty flask at an initial pressure of 0.500 atm and a temperature above 500 K. Ov

tia_tia [17]

Answer:

Kp = 0.81666

Explanation:

Pressure of PCl₅ = 0.500 atm

Considering the ICE table for the equilibrium as:

PCl₅ (g) ⇔ PCl₃ (g) + Cl₂ (g)

t = o 0.500

t = eq -x x x

--------------------------------------------- --------------------------

Moles at eq: 0.500-x x x

Given the pressure of PCl₅ at equilibrium = 0.150 atm

Thus, 0.500 - x = 0.150

x = 0.350 atm

The expression for the equilibrium constant is:

$K_p=\frac {P_{PCl_3}P_{[Cl_2}}{P_{PCl_5}}$

So,

$K_p=\frac{x^2}{0.500-x}$

x = 0.350 atm

Thus,

$K_p=\frac{{0.350}^2}{0.500-0.350}$

7 0

3 years ago

The nuclear binding energy of one nitrogen-14 atom is (BLANK) x 10A J. Round to 3

d1i1m1o1n [39]

mass defect = mass of constituents - mass of atom

N has 7p and 9n

proton mass ~ 1.00728 amu

neutron mass ~ 1.00866 amu

electron mass ~ 0.000549 amu

Nitrogen mass ~ 14.003074 amu

mass defect = (7*1.00728)-(7*1.00866)-(7*0.000549) - 14.003074

= 0.11235amu

convert to energy, the binding energy = 1.68x10^-11 J

5 0

3 years ago

Is water used to test cheese for protein

nekit [7.7K]

Answer:

yes

Explanation:

because it help the body hydrate

7 0

3 years ago

A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path

Igoryamba

Explanation:

For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

R = 8.314 J/(K mol), T = 298 K ,

$V_{2}$ = 8.34 L, $V_{1}$ = 2.67 L

Now, putting the given values into the above formula as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494$

= -2818.68 J

Hence, work for path A is -2818.68 J.

For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure $P_{external}$ = 1.00 atm.

So, W = $-P_{external} \times \Delta V$

Now, putting the given values into the above formula as follows.

W = $-P_{external} \times \Delta V$

= $-1 atm \times (8.34 L - 2.67 L)$

= -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

W = $-\frac{101.33 J}{1 atm L} \times 5.67 atm L$

= -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

W = -574.54 J

Hence, work for path B is -574.54 J.

3 0

3 years ago