All categories

3 years ago

What is the % Hydrogen in glucose

Chemistry

1 answer:

sleet_krkn [62]3 years ago

5 0

Answer:

50%

Explanation:

Hydrogen makes up about 50% in glucose in atoms, in mass, 6.7 wise

You might be interested in

Why is it important to analyze data?

gayaneshka [121]

Answer:

It is important to analyze data to further understand what's going on. By analyzing data, you know more about what you are investigating. Without analyzing data, you may find it harder to figure something.

7 0

3 years ago

Mercury has a density of 13.6 grams/mL, what is this density in cg/L?

Firdavs [7]

It would be 4.6cgL not sure tho because I didint do that good in this

4 0

3 years ago

Consider the following equilibrium:

Ainat [17]

Answer: The equation which is wrong is $K_p=K_c(RT)^{-5}$

Explanation:

For the given reaction:

$4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)$

The expression for $K_c\text{ and }K_p$ is given by:

$K_c=\frac{1}{[O_2]^3}$

$K_p=\frac{1}{[O_2]^3}$

The concentration of solids are taken to be 1, only concentration of gases and liquid states are taken. The pressure of only gases are taken.

Relationship between $K_p\text{ and }K_c$ is given by the expression:

$K_p=K_c\times (RT)^{\Delta n_g}$

where,

$\Delta n_g$ = number of moles of gaseous products - number of moles of gaseous reactants

R = gas constant

T= temperature

For the above reaction,

$\Delta n_g$ = number of moles of gaseous products - number of moles of gaseous reactants = 0 - 3 = -3

Hence, the expression for $K_p$ is:

$K_p=K_c\times (RT)^{-3}$

Therefore, the equation which is wrong is $K_p=K_c(RT)^{-5}$

7 0

3 years ago

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For Iron:

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

3 years ago

How to answer all these questions?

Fudgin [204]

You have to do it to see the results

8 0

3 years ago