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4 years ago

HELP HELP HELP!!!!

2 answers:

5 0

Well metals are good conductors of heat and electricity because in a storm you don't wanna be by metal or holding metal in the air because the lighting can strike it at any moment, plus when we plug something in we see a metal piece going into the socket, and because we use pots and pans to bake and cook food so they would be good conductors of heat because if you take them out with your bare hands your hands will get Burt because its metal.

Hope that helped!

notka56 [123]4 years ago

4 0

Metals and metalloids :)))

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What is the final pressure (expressed in atm) of a 3.05 l system initially at 724 mm hg and 298 k, that is compressed to a final

krok68 [10]

Answer:

P2 = 778.05 mm Hg = 1.02 atm

Explanation:

We are to find the final pressure (expressed in atm) of a 3.05 liter system initially at 724 mm hg and 298 K which is compressed to a final volume of 2.60 liter at 273 K.

For this, we would use the equation:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

where P1 = 724 mm hg

V1 = 3.05 L

T1 = 298 K

P2 = ?

V2 = 2.6 L

T2 = 173 K

Substituting the given values in the equation to get:

$\frac{(724)(3.05)}{298} =\frac{P_2(2.6)}{173}$

P2 = 778.05 mm Hg = 1.02 atm

7 0

3 years ago

What is the pH for a solution that has an OH ion

Basile [38]

Assuming that you mean 10^-4 M then this would be basic and would have a pH of 10.

pOH = -log[OH].
So pOH = 4
pH=14-pOH
pH = 10

3 0

3 years ago

Suppose that you add 29.2 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the adde

enyata [817]

Answer:

Suppose that you add 29.2 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 2.78 oC compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound

Explanation:

The mass of nonvolatile solute added is ---- 29.2g

The mass of solvent benzene is ---- 0.250kg = 250g

The Kf value of benzene is ---- 5.12^oC/m.

Depression in the freezing point of the solution is --- 2.78^oC.

What is the molar mass of the unknown solute?

$The depression in freezing point = Kf * molality of the solution\\molality of the solution = \frac{mass of solute}{molar mass of solute}*\frac{1}{mass of solvent in kg}$

Substitute the given values in this formula to get the molar mass of unknown solvent:

$molality=\frac{29.2g}{M} * \frac{1}{0.250kg} \\depression in freezing point:\\2.78^oC=5.12^oC/m * \frac{29.2g}{M} * \frac{1}{0.250kg} \\\\=>M=5.12^oC/m * \frac{29.2g}{2.78^oC} * \frac{1}{0.250g} \\\\\\=>M=215.1g/mol$

Hence, the molar mass of unknown solute is --- 215g/mol.

5 0

3 years ago

How many bonds can a carbon atom form? A. 3 O B. 4 C. 2. O D. 1

IrinaK [193]

Answer:

B)4

Explanation:

5 0

3 years ago

Assume a scuba diver is working at a depth where the total pressure is 5.0 atm (about 50 m below the surface). What mole fractio

qwelly [4]

Answer: 0.042

Explanation:

The partial pressure of a gas is given by Raoult's law, which is:

$p=p_T\times \chi$

where,

$p_A$ = partial pressure of oxygen = 0.210 atm

$p_T$ = total pressure = 5.0 atm

$\chi_A$ = mole fraction of oxygen = ?

Putting values in above equation, we get:

$0.210=5.0\times \chi$

$\chi =0.042$

Hence, the mole fraction of oxygen necessary in order for the partial pressure of oxygen in the gas mixture the diver breathes to be 0.210 atm is 0.042.

4 0

3 years ago