Which is a density independent factor limiting popultion growth

Balance this redox reaction occurring in acidic media:

Lelu [443]

Add 7 water atom to the right hand side to adjust the quantity of oxygen. Increase Cr(+3) by two to adjust the quantity of Cr. Duplicate Cl-by two to adjust the quantity of chlorine molecules.
Cr2O7[2-](aq) +2 Cl[-](aq) < - >2 Cr[3+] (aq) + Cl2(g)+7H2O
Presently adjust that charges.
you have - 4 charges on the left hand side, while +18 charges on the right hand side, there for include 14H+ the left hand side to adjust the charges
Cr2O7[2-](aq) +2 Cl[-](aq)+14H+ < - >2 Cr[3+] (aq) + Cl2(g)+7H2O
take note of that the oxidation number of hydrogen in water is +1

3 0

3 years ago

Question 2

Thepotemich [5.8K]

Answer:

1-Pentene

Explanation:

If we look at all the options listed, we will notice that the rate of reaction of bromine with each one differs significantly.

For 1-pentene, addition of bromine across the double bond is a relatively fast process. It is usually used as a test for unsaturation. Bromine water is easily decolorized by alkenes.

Cyclohexane, heptane are alkanes. They can only react with chlorine in the presence of sunlight. This is a substitution reaction. It does not occur easily. A certain quantum of light is required for the reaction to occur.

For benzene, bromine can only react with it by electrophilic substitution in which the benzene ring is retained. A Lewis acid is often required for the reaction to occur and it doesn't occur easily.

3 0

3 years ago

Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a cat

zlopas [31]

Answer:

48.8%

Explanation:

The reaction has a 1:1 mole ratio so;

Number of moles of benzoic acid reacted = mass/molar mass = 3.8 g/122.12 g/mol = 0.03 moles

So;

0.03 moles of methyl benzoate is formed in the reaction

Mass of methyl benzoate formed = 0.03 moles * 136.15 g/mol = 4.1 g

percent yield = actual yield/theoretical yield * 100/1

percent yield = 2.0 g/4.1 g * 100 = 48.8%

4 0

3 years ago

Balance the following chemical equation, then answer the following question.

julsineya [31]

Molar mass:

O2 = 31.99 g/mol
C8H18 = 144.22 g/mol

2 C8H18(g) + 25 O2(g) = 16 CO2(g) + 18 H2O(g)

2 x 144.22 g --------------- 25 x 31.99 g
10.0 g ----------------------?? ( mass of O2)

10.0 x 25 x 31.99 / 2 x 144.22 =

7997.5 / 288.44 => 27.72 g of O2

hope this helps!

3 0

3 years ago

During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200.

Travka [436]

Answer: The mass of HCl present in 500 mL of acid solution is 36.5 grams

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL$

Putting values in above equation, we get:

$1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

$2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g$

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

4 0

3 years ago