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3 years ago

7

Sailin' Steve likes to test the buoyancy of different objects. He threw a spherical object into a fluid that has a greater densi

ty than the object does. He also threw a square object into the fluid that has a greater density than the fluid.

1 answer:

Hunter-Best [27]3 years ago

7 0

The spherical object will float and the square object will sink to the bottom of the fluid as it has greater density. :)

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What is the molarity of a solution containing 23 g of NaCl in 500 mL of solution?

trapecia [35]

molarity of a solution means mols per liter.

First, you need to convert 23 grams on NaCl into mols. 23g divided by molar mass (58.44g/mol) which gives you .394 mols.

Now, you need to convert 500ml to L which moves the decimal three places to the left, giving you .500L of solution.

Finally, divide the mols over solution to get .787M

7 0

3 years ago

How can one kg of iron melt more ice than 1 kg lead at 100 °C

Vanyuwa [196]

Answer:

Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)

Explanation:

The given parameters are;

The metals provided to melt the ice and their temperature includes;

One kg (1000 g) of iron;

Specific heat capacity = 0.444 J/(g·°C)

Temperature = 100°C

1 kg (1000 g) of lead

Specific heat capacity = 0.160 J/(g·°C)

Temperature = 100°C

Therefore, the heat provided to the ice of mass m, and latent heat of 334 J/g at 0°C by the metals are as follows;

For iron, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ $_{iron}$ = Heat obtained from the iron by the ice

ΔQ $_{iron}$ = 0.444 m × 1000 × (100 - 0) = 44400 J

Heat absorbed by the ice for melting, H $_l$ = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H $_l$ = Mass of ice × Latent heat of ice

H $_l$ = Mass of ice × 334 J/g = 44400 J

Mass of ice melted by the iron = 44400 J/334 (J/g) ≈ 132.9 g

Mass of ice melted by the iron ≈ 132.9 g

For lead, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ $_{lead}$ = Heat obtained from the iron by the ice

ΔQ $_{lead}$ = 0.160 m × 1000 × (100 - 0) = 16000 J

Heat absorbed by the ice for melting, H $_l$ = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H $_l$ = Mass of ice × Latent heat of ice

H $_l$ = Mass of ice × 334 J/g = 16000 J

Mass of ice melted by the lead = 16000 J/334 (J/g) ≈ 47.9 g

Mass of ice melted by the lead ≈ 47.9 g

Therefore, mass of ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.

3 0

3 years ago

Calculate the ph of sweater which has a hydrogen ion concentration of 1 times 10 to the power of 8

telo118 [61]

Answer:

The pH of the sweater containing Hydrogen ion concentration

$[H^{+}]=1\times 10^{-8}$ is

<u>8</u>

<u></u>

Explanation:

pH = It is the negative logarithm of activity (concentration) of hydrogen ions.

pH = -log([H+])

Now, In the question the concentration of [H+] ions is :

$[H^{+}]=1\times 10^{-8}$

$pH = -log(1\times 10^{-8})$

use the relation:

$log(10^{-a})=a$

$pH= -(-8)$

pH = 8

Note : <em><u> 1 times 10 to the power of 8 must be" 1 times 10 to the power of -8"</u></em>

If the concentration is

$[H^{+}]=1\times 10^{8}$

Then pH = -8 , which is not possible . So in that case the pH calculation is by other method

5 0

3 years ago

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

<u>For process 2 :</u>

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago

What element has the electron configuration 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2?

Marina CMI [18]

<span>6s²4f¹⁴5d¹⁰6p²
6 shows that the element is in the 6 period,
6p² shows that the element is in the 14th group. (1 and 2 groups have s -electrons as last ones, 13 group has s²p¹, and 14 group has s²p²)
The element is Pb.

</span>

3 0

3 years ago

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