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3 years ago

Develop a definition of chemical change using atoms and molecules in your definition. (please help)

Chemistry

1 answer:

Arisa [49]3 years ago

8 0

Answer:

a chemical change is a substance change in to a new substance

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Calculate the boiling point of a solution of 500.0 g of ethylene glycol (c2h6o2) dissolved in 500.0 g of water. kf = 1.86Â°c/m a

Bingel [31]

Answer:The boiling point of the solution is 108° C.

Explanation:

Boiling point of pure water=T= $100^oC$

Boiling point of water after addition of 500 g of ethylene glycol= $T_f$

Mass of water = 500g = 0.5 kg (1000 g = 1 kg)

$\Delta T_f=K_b\times \frac{\text{Mass of ethlyene glycol}}{\text{Molar mass of ethylene glycol}}$

$\Delta T_f=0.512^oC/m\times \frac{500 g}{62.07 g/mol\times 0.5 kg}$

$\Delta T_f=8.24 ^oC$

$\Delta T_f=T_f-T$

$8.24^oC=T_f-100^oC$

$T_f=108.24^oC$

The boiling point of the solution is 108° C.

7 0

3 years ago

Write a balenced chemical equation for the reaction: include abbreviation for the physical states-

Nuetrik [128]

Answer:

2Na(s) + 2H2O(l) ------> 2NaOH(aq) + H2(g)

Explanation:

5 0

3 years ago

The exhaust gas from an automobile contains 3% by volume of carbon monoxide (CO). Express this concentration in mg/m3 at 25oC an

ser-zykov [4K]

Answer:

$24540\frac{mg}{m^3}$

Explanation:

Hello,

In this case, since the 3% by volume is represented as:

$\frac{3L\ CO}{L\ gas}$

By using the ideal gas equation we compute the density of CO:

$\rho =\frac{MP}{RT} =\frac{28g/mol*1atm}{0.082\frac{atm*L}{mol*K}*298K}= 0.818g/L$

Then we apply the conversion factors as follows:

$=\frac{3L\ CO}{100L\ gas}*\frac{0.818g\ CO}{1L\ CO} *\frac{1000mg\ CO}{1g\ CO} *\frac{1000L\ gas}{1m^3\ gas} \\\\=24540\frac{mg}{m^3}$

Regards.

4 0

4 years ago

I NEED THIS ASAP TODAY PLEASSSSEEEEEE

alukav5142 [94]

Option C, mass would be same. Only the gravitational pull will be different

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

4 years ago