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3 years ago

50 points in total i believe! Why does chocolate melt when it gets warm?

Chemistry

1 answer:

CaHeK987 [17]3 years ago

4 0

Answer:

Because it is made of a hardened liquid that is extremely sensitive to heat

Explanation:

Brainliest?

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Would magnesium prefer to lose or gain electrons

slega [8]

Umm what kind of question is that but i think gain lol

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3 years ago

What is the correct name for SO3?

ElenaW [278]

Sulfur Trioxide is the correct name for SO3

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3 years ago

The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

3 years ago

Elements within the same group of the periodic table behave similarly because they have the same number of what

Darina [25.2K]

Same number of ions
Yeeeeeeeeee

4 0

3 years ago

Read 2 more answers

If 2.03 g of oxygen react with carbon monoxide, how many grams of CO2 are formed?

mamaluj [8]

Answer:

The suitable equation for this reaction is

2CO + O₂ -----> 2CO₂

Here, we are given that we have 2 grams of O₂

From the equation, we can see that 2 * Moles of O₂ = Moles of CO₂

Moles of O₂:

2/32 = 1/16 moles

Therefore, the number of moles of CO₂ is twice the moles of O₂

Moles of CO₂ = 2 * 1/16

Moles of CO₂ formed = 1/8 moles

Mass of CO₂ formed = Molar mass of CO₂ * Moles of CO₂

Mass of CO₂ formed = 44 * 1/8

Mass of CO₂ formed = 5.5 grams

Hence, option B is correct

Kindly Mark Brainliest, Thanks!!!

8 0

3 years ago