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4 years ago

How many carbon are contained 25.00 g of sugar (C12H22O11)

1 answer:

7 0

342 gm sugar = 1 mol of sugar = 12 moles of C. Therefore 25 gm of sugar will form = 25x12/342 = 0.877 moles of C. Therefore number of carbon atoms = number of moles x 6.02x 10^23 = 0.877x 6.02x 10^23 = 5.28x10^23.

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What happens when chemical bonds break and new bonds form

zheka24 [161]

Answer:

Chemical reactions make and break the chemical bonds between molecules, resulting in new materials as the products of the chemical reaction.

Explanation:

Breaking chemical bonds absorbs energy, while making new bonds releases energy, with the overall chemical reaction being endothermic or exothermic.

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3 years ago

Which comparison accurately states why phylogenetic trees based on molecular data can be more informative than those based on mo

Romashka [77]

Molecular data is more informative than morphology because this examines the DNA sequence and protein composition. This then provide a more comprehensive detail on what species are more related and distant with one another. .
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3 years ago

Which equation has x = 5 as the solution?

Natali5045456 [20]

Answer:

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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

Given the following equation: 2 K3N ---> 6 K + N2 How many moles of N2 can be produced by letting 12.00 moles of K3N react?

myrzilka [38]

The answer will be 12 nitrogen or N2 will be produced because if you changed the coefficient to 12 on the reactant side and distribute, the nitrogen would be 12 when distributed and what happens on one side has to equal to the other side, which is the product side.

8 0

3 years ago