Describe, on a molecular level, how you would expect these lipids to behave in water.

3. The alcohol in "gasohol" burns according to the following equation. C2H6O + 3 O2 --------> 2 CO2 + 3 H2O (a) If 25 moles o

Vanyuwa [196]

For all question, all you need to use is the mole-mole ratio.

a) 25 moles C2H6O (3 moles O2/ 1 mol C2H6O)= 75 moles O2

b) 30 moles O2 (1 moles C2H6O/ 3 moles O2)= 10 moles C2H6O

c) 23 moles CO2 (3 moles O2/ 2 moles CO2) = 34.5 moles O2

d) 41 moles H2O ( 1 moles C2H6O/ 3 moles H2O= 13.7 moles C2H6O

7 0

3 years ago

How many grams of glucose, C6H12O6, in 2.47 mole?

statuscvo [17]

First, we need to find the atomic mass of $C_{6}H_{12}O_{6}$ .

According to the periodic table:
The atomic mass of Carbon = C = 12.01
The atomic mass of Hydrogen = H = 1.008
The atomic mass of Oxygen = O = 16

As there are 6 Carbons, 12 Hydrogens and 6 Oxygens, therefore:
The molar mass of   $C_{6}H_{12}O_{6}$ = 6 * 12.01 + 12 * 1.008 + 6 * 16

The molar mass of   $C_{6}H_{12}O_{6}$ = 180.156 grams/mole

Now that we have the molar mass of   $C_{6}H_{12}O_{6}$ , we can find the grams of glucose by using:

mass(of glucose in grams) = moles(of glucose given in moles) * molar mass(in grams/mole)

Therefore,
mass(of glucose in grams) = 2.47 * 180.156
mass(of glucose in grams = 444.99 grams

Ans: Mass of glucose in grams in 2.47 moles = 444.99 grams

-i

6 0

3 years ago

Read 2 more answers

423.3 mL of gas is at 49.2 C. It is compressed to a volume of 79 mL. What is the new temperture. Express your answer in Kelvin.

zzz [600]

Answer: 60.1K

Explanation:

Initial volume of gas V1 = 423.3mL

Initial temperature T1 = 49.2°C

Convert temperature in Celsius to Kelvin

( 49.2°C + 273 = 322.2K)

Final temperature T2 = ?

Final volume V2 = 79mL

According to Charle's law, the volume of a fixed mass of a gas is directly proportional to the temperature.

Mathematically, Charles' Law is expressed as: V1/T1 = V2/T2

423.3mL/322.2 = 79mL/T2

To get the value of T2, cross multiply

423.3mL x T2 = 322.2K x 79mL

423.3mL x T2 = 25453.8

T2 = (25453.8/423.3mL)

T2 = 60.1K

Thus, the new temperature of the gas is 60.1K

8 0

3 years ago

A student needs to prepare 100. mL of 0.612 M Cu(NO3)2 solution. What mass, in grams, of copper(II) nitrate should the student u

Temka [501]

Answer: 11.5 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution

$Molarity=\frac{n\times 1000}{V_s}$

where,

Morality = 0.612 M

n= moles of solute

$V_s$ = volume of solution in ml = 100 ml

Now put all the given values in the formula of molarity, we get

$0.612=\frac{n\times 1000}{100ml}$

$n=0.0612moles$

$Mass={\text {moles of solute }}{\times {\text {molar mass}}=0.0612moles\times 187.56g/mol=11.5g$

Therefore, the mass of copper (II)nitrate required is 11.5 grams

3 0

3 years ago

Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: spe

Murljashka [212]

Answer:

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

Heat required to raise the temperature of ice from -20 °C to 0 °C

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 $\frac{J}{g*C}$ and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 $\frac{J}{g*C}$ *20 C and solving: Q=420 J

Heat required to convert 0 °C ice to 0 °C water

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g* $6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}$ = 3.333 kJ= 3,333 J (being kJ=1,000 J)

Heat required to raise the temperature of water from 0 °C to 60 °C

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 $\frac{J}{g*C}$ and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 $\frac{J}{g*C}$ *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

3 0

3 years ago