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2 years ago

What is included in each square (element) on the Periodic Table? I'm looking for four things

Chemistry

1 answer:

Veseljchak [2.6K]2 years ago

6 0

Answer:

name of element, symbol, atomic number, relative atomic mass (atomic weight)

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(02.02 LC) The two main types of weathering are Select one:

Firlakuza [10]

Answer:

Explanation:

4 0

2 years ago

Adding more than one equivalent of HCl to pent-1-yne will lead to which product:______.

Ket [755]

Answer:

c. 2,2-dichloropentane.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to firstly draw the structure of the reactant, pent-1-yne:

$CH\equiv C-CH_2-CH_2-CH_2$

Now, we infer the halogen is added to the carbon atom with the most carbon atoms next to it, in this case, carbon #2, in order to write the following product:

$CH\equiv C-CH_2-CH_2-CH_2+2HCl\rightarrow CH_3- CCl_2-CH_2-CH_2-CH_2$

Whose name is 2,2-dichloropentane.

Regards!

8 0

2 years ago

What is the name of Earth’s natural satellite?

Dafna11 [192]

Answer:

the moon is the earths natural satellite.

Explanation:

A natural satellite is in the most common usage, an astronomical body that orbits a planet, dwarf planet, or small Solar System body (or sometimes another natural satellite). Natural satellites are often colloquially referred to as moons, a derivation from the Moon of Earth.

4 0

2 years ago

The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k

Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

$ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

We have given in the question

$P_1=102\ mmHg$

$T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol$

And $R$ is the Universal Gas Constant.

$R=0.008 314 kJ/Kmol$

$ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165$

Taking inverse log both side we get,

$\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg$

8 0

3 years ago

To what Celsius temperature must 67.0 mL of krypton gas at 18.0°C be changed so the volume will triple? Assume the pressure and

Mice21 [21]

Answer: 600°C

Explanation:

This reaction is explained by Charles' law as the pressure is constant.

From the question, we obtained:

V1 = 67mL

T1 = 18°C = 18 +273 = 291K

V2 = 3V1 ( Vol is tripled) = 3x67 = 201mL

T2 =?

Applying the Charles' law,

V1 /T1 = V2 /T2

67/291 = 201 / T2

Cross multiply to express in linear form.

67xT2 = 291x201

Divide both side by 67, we have:

T2 = (291x201) /67

T2 = 873K

Converting to Celsius temperature, we have

T°C = K — 273

T°C = 873 — 273 = 600°C

8 0

3 years ago