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3 years ago

G- how many grams of f2 gas are there in a 5.00-l cylinder at 4.00 × 103 mm hg and 23°c?

1 answer:

6 0

<span>PV = nRT (4000 Torr)(5 L) = n (62.4 Torr-L/mol-K)(296K) n = 1.08 moles 28 g/mol, 1.08 moles = 30.3 grams your answer is C.30.3 g</span>

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Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,

8_murik_8 [283]

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

6 0

3 years ago

What event started the universe

olya-2409 [2.1K]

I believe the Big Bang

8 0

4 years ago

Help asap - chemistry, will mark brainlyest

melisa1 [442]

Answer:

Error percentage = 1.326% , 1.35% , 5.37%

density = mass / volume

For first sample:

density = 37.22/11.5 = 3.2365 g/ml

then percentage error = 100% - (3.2365/3.28 ) * 100 = 1.326%

For second sample:

density = 37.21/11.5 = 3.2356 g/ml

then percentage error = 100% - ( 3.2356 / 3.28 ) *100 = 1.35%

For third sample:

density = 37.25/12 = 3.104 g/ml

then percentage error = 100% - ( 3.104 / 3.28 ) *100 = 5.37%

3 0

3 years ago

How much heat will be absorbed by a 63.1 g piece of aluminum (specific heat = 0.930 J/g･°C) as it changes temperature from 23.0°

cluponka [151]

Answer:

$Q=2582J=2.58kJ$

Explanation:

Hello,

In this case, for us to compute the absorbed heat, we apply the following equation:

$Q=m_{Al}Cp_{Al}(T_2-T_1)$

Whereas we use the mass, specific heat and temperature change for the piece of aluminium, thus, we obtain:

$Q=63.1g*0.930\frac{J}{g*\°C}*(67.0-23.0)\°C\\ \\Q=2582J=2.58kJ$

It is positive as the heat is entering, therefore the temperature raises.

Best regards.

4 0

3 years ago

Karen and nbsp; measures the initial volume of water as 12.5ml and nbsp; she then adds a 2.3 gram rubber stopper and the volume

Ivenika [448]

Answer:

8.3mL

Explanation:

According to the question, the initial volume of a measured amount of water is 12.5mL. Karen then adds a rubber stopper that weighs 2.3grams. This increases the volume of the water.

The increase in the volume of the water is attributed to the rubber stopper. Hence, to calculate the volume of the rubber stopper, we subtract the initial volume of the water from the final volume. i.e.

Volume of added rubber stopper = final volume of water - initial volume of water

= 20.8 - 12.5

= 8.3mL

Hence, the volume of the rubber stopper is 8.3mL

8 0

4 years ago