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2 years ago

14

You add two substances together in a sealed bag and a chemical reaction happens. The mass before the reaction was 167 g. What wa

s the mass after the reaction? Why?

1 answer:

SpyIntel [72]2 years ago

4 0

167g. Because the law of conservation of mass says that matter can not be created or destoryed, in a chemical reaction.

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When do plants use the process of cellular respiration

d1i1m1o1n [39]

Answer:

So the answer is every day

Explanation:

Plants respire at all times of the day and night because their cells need a constant energy source to stay alive.

8 0

3 years ago

∆G° for the process benzene (l) benzene (g) is 3.7 Kj/mol at 60 °C , calculate the vapor pressure of benzene at 60 °C [R=0.0821

Alex787 [66]

Answer:

4) 0.26 atm

Explanation:

In the process:

Benzene(l) → Benzene(g)

ΔG° for this process is:

ΔG° = -RT ln Q

<em>Where Q = P(Benzene(g)) / P°benzene(l) P° = 1atm</em>

ΔG° = 3700J/mol = -8.314J/molK * (60°C + 273.15) ln P(benzene) / 1atm

1.336 = ln P(benzene) / 1atm

0.26atm = P(benzene)

Right answer is:

<h3>4) 0.26 atm </h3><h3 />

6 0

2 years ago

How to calculate the frequency of an ion

sergij07 [2.7K]

solution:

You need to find the frequency, and they have already given you the wavelength. And since you already know the speed of light, you can use formula (2) to answer this problem. Remember to convert the nano meters to meters because the speed of light is in meters. $(1 nm = 1.0 x 10^{-9} m)$

$v=c\lambda \\v= (3.00 \times 10^8 m/s)/(6.9\times 10^-7 m) \\v = 4.35 \times 10^{14} s^{-1}$

8 0

3 years ago

Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of

Brut [27]

<u>Answer:</u> The smallest temperature change is shown by water.

<u>Explanation:</u>

To calculate the heat absorbed or released, we use the equation:

$q=mC\times \Delta T$ ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

$\Delta T$ = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

<u>Option a:</u> 50.0 g Fe, $C_{Fe}=0.449J/g^oC$

We are given:

$m=50.0g\\C_{Fe}=0.449J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC$

Change in temperature = 8.99°C

<u>Option b:</u> 50.0 g water, $C_{water}=4.18J/g^oC$

We are given:

$m=50.0g\\C_{water}=4.18J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC$

Change in temperature = 0.96°C

<u>Option b:</u> 25.0 g Pb, $C_{Pb}=0.128J/g^oC$

We are given:

$m=50.0g\\C_{Pb}=0.128J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC$

Change in temperature = 62.5°C

<u>Option d:</u> 25.0 g Ag, $C_{Ag}=0.235J/g^oC$

We are given:

$m=25.0g\\C_{Ag}=0.235J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC$

Change in temperature = 34.04°C

<u>Option e:</u> 25.0 g granite, $C_{granite}=0.79J/g^oC$

We are given:

$m=25.0g\\C_{Fe}=0.79J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC$

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

5 0

3 years ago

Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should as

UNO [17]

Answer: 0.100 m $K_2SO_4$

Explanation:

Elevation in boiling point is given by:

$\Delta T_b=i\times K_b\times m$

$\Delta T_b=T_b-T_b^0$ = Elevation in boiling point

i= vant hoff factor

$K_b$ = boiling point constant

m= molality

1. For 0.100 m $K_2SO_{4}$

$K_2SO_4\rightarrow 2K^{+}+SO_4^{2-}$

, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be $3\times 0.100=0.300$

2. For 0.100 m $LiNO_3$

$LiNO_3\rightarrow Li^{+}+NO_3^{-}$

, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be $2\times 0.100=0.200$

3. For 0.200 m $C_2H_8O_3$

, i= 1 as it is a non electrolyte and does not dissociate, concentration of ions will be $1\times 0.200=0.200$

4. For 0.060 m $Na_3PO_4$

$Na_3PO_4\rightarrow 3Na^{+}+PO_4^{3-}$

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be $4\times 0.060=0.24m$

Thus as concentration of solute is highest for $K_2SO_4$ , the elevation in boiling point is highest and thus has the highest boiling point.

3 0

2 years ago