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galben [10]
3 years ago
10

Suppose you see a crescent Moon. How do you know if it is waxing or waning?

Chemistry
1 answer:
nasty-shy [4]3 years ago
5 0

Answer:

waxing is in the evening and waning is in the morning if you cant see the moon in the evening it is waning if you can see it in the evening it is waxing

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By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask
tangare [24]

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = V_1=11.00 mL

Concentration of stock solution = M_1=0.823 M

Volume of stock solution after dilution = V_2=50.00 mL

Concentration of stock solution after dilution = M_2=?

M_1V_1=M_2V_2 ( dilution )

M_2=\frac{0.823 M\times 11.00 mL}{50 ,00 mL}=0.18106 M

0.18106 M is the molarity of the resulting solution.

2)

Molarity of the solution is the moles of compound in 1 Liter solutions.

Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100  L ( 1 mL=0.001 L)

Molarity=\frac{13.0 g}{158 g/mol\times 0.100 L}=0.823 mol/L

0.823 Molar is the molarity of the solution.

6 0
3 years ago
A 2.5 g sample of french fries is placed in a calorimeter with 500.0 g of water at an initial temperature of 21 °C. After combus
SIZIF [17.4K]
Q=m°C<span>ΔT
=(500g) x (1 cal/g.</span>°C) x (48°C-21°C) = 13500 cal
13500 cal / 1000 = 13.5 kcal

<span>"What is the caloric value (kcal/g) of the french fries?"
13.5 kcal/ 2.5 g = 5.4 kcal/g</span>
8 0
4 years ago
A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2
sukhopar [10]

Answer:

a) \triangle G^{0} = 7.31 kJ/mol

b) K_{-1} = 0.0594 m^{-1} s^{-1}

Explanation:

Equation of reaction:

                                     2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)

Initial pressure                  3              1              0

Pressure change             2P           1P             2P

Total pressure = (3-2P) + (1-P) + (2P)

Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }

K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}

Standard free energy:

\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol

b) value of k−1 at 27 °C, i.e. 300K

K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}

K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}

K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2}  }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}

6 0
4 years ago
Study of gene structure and function
gregori [183]

I believe the answer would be b

7 0
3 years ago
How many moles of carbon are in 3.7 moles of C 8 H11 NO2
Dafna1 [17]
In every molecule of C_{8} H_{11}NO_2 there is 8 atoms of Carbon.

IF we have 3.7 moles of C_{8} H_{11}NO_2 to find the number of moles of Carbon, just multiply by 8

3.7 * 8 = 29.6 mol Carbon
6 0
3 years ago
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