Ask question

Ask question

All categories

11111nata11111 [884]

3 years ago

15

Heating glucose, a monosaccharide sugar, in the presence of excess oxygen produces carbon dioxide gas and water vapor. Balance t

his reaction. ___C6H12O6(s) + ___O2(g) → ___CO2(g) + ___H2O(g)

1 answer:

lozanna [386]3 years ago

6 0

Answer:

The balance reaction is C6H12O6+6O2=6CO2+6H2O

Explanation:

The reaction mentioned above is the reaction of cellular respiration.In this reaction glucose molecule reacts with oxygen to generate carbon dioxide,water and energy in form of ATP.

In the left side there are 6 carbon atoms,12 hydrogen atoms and 18 oxygen atoms and the same number of carbon,hydrogen and oxygen is present in the right side of the reaction mentioned above.Thus the reaction can be balanced.

You might be interested in

Drag each tile to the correct box.

77julia77 [94]

this is my attachment answer hope it's helpful to you

3 0

3 years ago

Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3

vichka [17]

Answer:

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

Explanation:

Elevation in boiling is given by :

$\Delta T_b=i\times k_b\times m$

Where :

i = van't Hoff factor

$k_b$ = Molal Elevation constant of solvent

m = molaity of the solution

1) 0.19 m of $AgNO_3$

$AgNO_3\rightarrow Ag^++NO_3^{-}$

i = 2 (electrolyte)

Molality of the solution = 0.19

Elevation is boiling point of solution:

$\Delta T_b=2\times k_b\times 0.19 m$

$\Delta T_b=0.38 m\times k_b$

2) 0.17 m of $CrSO_4$

$CrSO_4\rightarrow Cr^{2+}+SO_4^{2-}$

i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

$\Delta T_b=2\times k_b\times 0.17 m$

$\Delta T_b=0.34 m\times k_b$

3) 0.13 m of $Mn(NO_3)_2$

$Mn(NO_3)_2\rightarrow Mn^{2+}+2NO_3^{-}$

i = 3 (electrolyte)

Molality of the solution = 0.13

Elevation is boiling point solution :

$\Delta T_b=3\times k_b\times 0.13 m$

$\Delta T_b=0.39 m\times k_b$

4) 0.31 m Sucrose (nonelectrolyte)

i = 1 ( non electrolyte)

Molality of the solution = 0.31 m

Elevation is boiling point solution :

$\Delta T_b=1\times k_b\times 0.31 m$

$\Delta T_b=0.31 m\times k_b$

Higher the value of elevation in temperature higher will be the boiling point of the solution .

The decreasing order of solution from highest boiling point to lowest boiling point is :

$0.39 m\times k_b>0.38 m\times k_b>0.34 m\times k_b>0.31 m\times k_b$

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

6 0

3 years ago

Why are yeast cells used in making bread?

tatiyna

Answer:

Yeast is used for the leavening of bread. Yeast uses the sugars and oxygen in dough to produce more yeast cells and carbon dioxide gas. The carbon dioxide makes the dough rise which gives the bread a light and spongy texture. Yeast also works on the gluten network.

Explanation:

5 0

2 years ago

How many moles of Ca(OH)2 are in 3.5kg of Ca(OH)2? Answer in units of mole

Kobotan [32]

First, you need to convert kg to g.
So, 1 kg =1000g.
3.5 x 1000 = 3500g Ca(OH)2

We need to know the molar mass of Ca(OH)2.
Ca= 40.08 g
O=2(15.999)
H=2(1.0079)

Add them all together and you get 74.0938 g.

Put it in the formula from mass to moles.

# of moles = grams Ca(OH)2 x 1 mol Ca(OH)2
--------------------
molar mass Ca(OH)2

3500 g Ca(OH)2 x 1 mol Ca(OH)2
---------------------
74.0938 g Ca(OH)2

So divide 1/74.0938 and multiply by 3500.

You will get about 47.24 moles Ca(OH)2.

Hope this helps! :)

7 0

3 years ago

Read 2 more answers

A soft silvery metal has two naturally occurring isotopes: mass 84.9118, accounting for 72.15% and mass 86.9092, accounting for

zloy xaker [14]

Answer: The atomic weight of the metal would be 85.47.

Explanation:

Mass of isotope 1 of metal = 84.9118

% abundance of isotope 1 of metal = 72.15% = $\frac{72.15}{100}$

Mass of isotope 2 of metal= 86.9092

% abundance of isotope 2 of metal = 27.85% = $\frac{27.85}{100}$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[(84.9118)\times \frac{72.15}{100})+(86.9092)\times \frac{27.85}{100}]]$

$A=85.47$

Therefore, the atomic weight of the metal would be 85.47.

6 0

3 years ago