GIVING BRAINLYYY

photoshop1234 [79]

Answer:

Me to I need help please

5 0

3 years ago

Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0

grandymaker [24]

The given question is incomplete. The complete question is:

Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis. Determine the mass of glucose (C6H1206) produced

Answer: 60.0 g of glucose

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $CO_2$

$\text{Number of moles}=\frac{88.0g}{44g/mol}=2.0moles$

b) moles of $H_2O$

$\text{Number of moles}=\frac{64.0g}{18g/mol}=3.5moles$

$6CO_2+6H_2O\rightarrow C_{6}H_{12}O_6+6O_2$

According to stoichiometry :

6 moles of $CO_2$ require = 6 moles of $H_2O$

Thus 2.0 moles of $CO_2$ require= $\frac{6}{6}\times 2.0=2.0moles$ of $H_2O$

Thus $CO_2$ is the limiting reagent as it limits the formation of product.

As 6 moles of $CO_2$ give = 1 moles of glucose

Thus 2.0 moles of $CO_2$ give = $\frac{1}{6}\times 2.0=0.33moles$ of glucose

Mass of glucose = $moles\times {\text {Molar mass}}=0.33moles\times 180g/mol=60g$

Thus 60.0 g of glucose will be produced from 88.0 g of carbon dioxide and 64.0 g of water

8 0

3 years ago

You’ll soon make two pieces of ice with the approximate shapes shown in the image. Both pieces of ice have the same mass and vol

Montano1993 [528]

Is there an image? I'll stay to help you.

3 0

4 years ago

At high temperatures, carbon reacts with O2 to produce CO as follows: C(s) O2(g) 2CO(g). When 0.350 mol of O2 and excess carbon

777dan777 [17]

Answer:

Value of $K_{c}$ is 0.090.

Explanation:

Initial molarity of $O_{2}$ = $\frac{0.350}{5.00}M$ = 0.0700 M

Construct an ICE table corresponding to the combustion reaction of carbon to determine $K_{c}$

$C(s)+O_{2}(g)\rightarrow 2CO(g)$

I (M): - 0.0700 0

C (M): - -x +2x

E (M): - 0.0700-x 2x

So, $K_{c}=\frac{[CO]^{2}}{[O_{2}]}$ , where [CO] and $[O_{2}]$ represents equilibrium concentration of CO and $O_{2}$ respectively.

Here, $[CO]=2x=0.060$

⇒x = 0.030

So, $[O_{2}]$ = 0.0700-x = (0.0700-0.030) = 0.040

Hence, $K_{c}=\frac{(0.060)^{2}}{0.040}=0.090$

4 0

4 years ago

93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0 C.what would be the volume of this dry gas at STP

ExtremeBDS [4]

Let us assume that the oxygen behaves as an ideal gas such that we can use the ideal gas equation to solve for the number of moles of O2.
PV = nRT ; n = PV/RT
Substituting the known values,
n = (0.930 atm)(93/1000 L) / (0.0821 L.atm/mol.K)(10 + 273.15K)
n = 3.72 x 10^-3 mols
At STP, the volume of each mol of gas is equal to 22.4 L.
volume = (3.73 x 10^-3 mols) x (22.4 L/1 mol)
volume = 0.0833 L or 83.34 mL

5 0

4 years ago