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Ivanshal [37]
3 years ago
5

A substance has a high melting point and conducts electricity in the liquid phase The is substance is

Chemistry
1 answer:
posledela3 years ago
3 0

Answer:

oxygen

Explanation:

?

You might be interested in
How many liters are in 3.45 moles of oxygen gas at STP?
just olya [345]

Answer:

22.4L of one mole of any gas

or you can use PV=nRT

3.45*22.4=77.28

Explanation:

5 0
3 years ago
A sample of gas occupies 9.0 mL at a pressure of 500.0 mm Hg. A new volume of the same sample is at a pressure of 750.0 mm Hg.
Anuta_ua [19.1K]
<h3>Answer:</h3>

                The New pressure (750 mmHg) is greater than the original pressure (500 mmHg) hence, the new volume (6.0 mL) is smaller than the original volume (9.0 mL).

<h3>Solution:</h3>

              According to Boyle's Law, " <em>The Volume of a given mass of gas at constant temperature is inversely proportional to the applied Pressure</em>". Mathematically, the initial and final states of gas are given as,

                                     P₁ V₁  =  P₂ V₂    ----------- (1)

Data Given;

                  P₁  =  500 mmHg

                  V₁  =  9.0 mL

                  P₂  =  750 mmHg

                  V₂  =  ??

Solving equation 1 for V₂,

                   V₂  =  P₁ V₁ / P₂

Putting values,

                   V₂  =  (500 mmHg × 9.0 mL) ÷ 750 mmHg

                   V₂  =  6.0 mL

<h3>Result:</h3>

            The New pressure (750 mmHg) is greater than the original pressure (500 mmHg) hence, the new volume (6.0 mL) is smaller than the original volume (9.0 mL).

4 0
4 years ago
Read 2 more answers
BY ANSWERING THIS QUESTION UR PUTTING IT ON UR MOM's LIFE THAT U WON'T STEAL MY POINTS.
Yakvenalex [24]

Answer:

T_2=-125.58\°C

Explanation:

Hello!

In this case, considering the Gay-Lussac's law which describes the pressure-temperature behavior as a directly proportional relationship by holding the volume as constant, we write:

\frac{T_1}{P_1} =\frac{T_2}{P_2}

Whereas solving for the final temperature T2, we get:

T_2=\frac{T_1P_2}{P_1}

Thus, we plug in the given data (temperature in Kelvins) to obtain:

T_2=\frac{(22+273.15)K*1.75atm}{3.50atm} \\\\T_2=147.58K-273.15\\\\T_2=-125.58\°C

Best regards!

3 0
3 years ago
I'm stuck on this assignment, +20 PTS and BRAINLIEST for step by step:
dolphi86 [110]

Answer:

2. 2.74 L

3. 488 K

4. 7.47 L

5. 38.6 L

6. 2.85 mol

7. 319 K

8. 3.43 kPa

Explanation:

Generally speaking, you want to convert units to SI units, but in this case, we are working with ratios.  This makes up for using the units that wouldn't appropriate elsewhere.

2.  Use the equation P₁V₁ = P₂V₂.  Solve for V₂.  

(3.05 L)(870 kPa) = (969 kPa)(V₂)  

V₂ = 2.74 L.  

3.  Use the equation V₁/T₁ = V₂/T₂. Solve for T₂.  

(3.32 L)/(360 K) = (4.50 L)/(T₂)  

T₂ = 488 K.

4.  Do the same as above, but for V₂.  

(5.10 L)/(-56°C) = V₂/(-82°C)  

V₂ = 7.47 L

5.  Use the equation V₁/n₁ = V₂/n₂.  Solve for V₂.  

(37.2 L)/(0.750 mol) = (V₂)/(0.778 mol)  

V₂ = 38.6 L

6.  Do the same as above, but for n₂.

(86.0 L)/(2.65 mol) = (92.5 L)(n₂)  

n₂ = 2.85 mol

7.  Use the equation P₁/T₁ = P₂/T₂.  Solve for T₂.  

(3.00 atm)/(390 K) = (2.45 atm)/(T₂)  

T₂ = 319 K

8.  Do the same as above, but for P₂.  

In this specific case, however you will need to convert units.  Since both temperatures don't have the same sign, the ratio won't come out right.  Convert to Kelvin.  Add 273.15 to the temperature in Celsius to convert to Kelvin -12.3°C = 260.85 K  25°C = 298.15 K.

(3.00 kPa)/(260.85 K) = P₂/(298.15 K)

P₂ = 3.43 kPa

There is a lot in here... If you are confused about something, let me know!

6 0
3 years ago
Lindsey is trying to gain credibility for her studies. She completed her experiment and discussed her finding with colleagues. H
noname [10]

Answer:

Option B:Publishing scientific journals

Explanation:

We are told that Lindsey is trying to gain credibility for her studies.

Since she completed her experiment and discussed her finding with colleagues, the most logical next step would be to publish scientific journals. This is because the other options given are not steps that should be taken because she has completed the research and therefore has no need to speak at a conference next nor even create new charts which they must have done during the research. No need for her to make sure the topic is popular.

Option B is correct

8 0
3 years ago
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