CH3CHClCH(CH3)CH2CH2CH2CH2Br name the molecule iupac rules please

As an ice cube melts on a table top it transforms from a solid bank with a defined shape to a puddle with a much greater ability

jeka94

In a solid state, the molecules have the least amount of energy. They just stick close together and vibrate in place. As the molecules gain more energy, they are able to move around more freely. In the liquid state, the molecules have enough energy to sort of tumble over each other.

5 0

3 years ago

Two types of organic reactions are

shepuryov [24]

2) deposition and saponification

3 0

3 years ago

3.

Dovator [93]

Answer : The number of grams of calcium perchlorate is, 0.00253 grams.

Explanation :

Molar mass of calcium perchlorate = 238.9 g/mol

As, $6.022\times 10^{23}$ formula units present in 238.9 g of calcium perchlorate

So, $6.37\times 10^{18}$ formula units present in $\frac{6.37\times 10^{18}}{6.022\times 10^{23}}\times 238.9=0.00253g$ of calcium perchlorate

Therefore, the number of grams of calcium perchlorate is, 0.00253 grams.

6 0

3 years ago

If 24.3 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over?2 NO (g) + O₂ (g) → 2

zhuklara [117]

Explanation:

2 NO (g) + O₂ (g) ----> 2 NO₂ (g)

24.3 g of NO are reacting with 13.8 g of O₂. First we can convert the mass of theses samples into moles using their molar masses.

molar mass of O = 16.00 g/mol

molar mass of N = 14.01 g/mol

molar mass of NO = 16.00 g/mol + 14.01 g/mol

molar mass of NO = 30.01 g/mol

molar mass of O₂ = 2 * 16.00 g/mol

molar mass of O₂ = 32.00 g/mol

moles of NO = 24.3 g * 1 mol/(30.01 g)

moles of NO = 0.810 moles

moles of O₂ = 13.8 g * 1 mol/(32.00 g)

moles of O₂ = 0.431 moles

Now, to determine the limiting reactant or the excess reactant we can find the number of moles of O₂ that will react with 0.810 moles of NO and the number of moles of NO that will react with 0.431 moles of O₂.

According to the coefficients of the reaction 2 moles of NO will react with 1 mol of O₂. Let's use that relationship to find the limiting reagent.

2 moles of NO = 1 mol of O₂

moles of O₂ = 0.810 moles of NO * 1 mol of O₂/(2 moles of NO)

moles of O₂ = 0.405 moles

moles of NO = 0.431 moles of O₂ * 2 moles of NO/(1 mol of O₂)

moles of NO = 0.862 moles

We found that we need 0.405 moles of O₂ to completely react with 0.810 moles of NO. Or, we need 0.862 moles of NO to completely react with ours 0.431 moles of NO.

We can say that NO is limiting our reaction and O₂ is in excess.

Only 0.405 moles of O₂ will react with 0.810 moles of NO. But we had 0.431 moles of it. Let's find the excess.

Excess of O₂ = 0.431 moles - 0.405 moles

Excess of O₂ = 0.026 moles

Answer: 0.026 moles is the number of moles of oxygen that will be left over.

4 0

1 year ago

1. Perform calculations to determine the amount of 6.00x10-5 M stock solution needed to prepare 20.00 mL of 2.00x10-5 M dye solu

zhannawk [14.2K]

Answer:

1a. 6.70 ml of stock dye solution is required

1b. 10.0 ml of stock dye solution is required

1c. 4.00 ml of stock dye solution is required

Explanation:

1a. Using m₁v₁ = m₂v

6.00 * 10⁻⁵ * v₁ = 20.0 * 2.00 * 10⁻⁵

v₁ = 4 * 10⁻⁴/6.00 * 10⁻⁵

v₁ = 6.70 mL of stock solution

Therefore, 6.70 ml of stock dye solution is required

b. Using m₁v₁ = m₂v

2.00 * 10⁻⁵ * v₁ = 20.0 * 1.00 * 10⁻⁵

v₁ = 2 * 10⁻⁴/2.00 * 10⁻⁵

v₁ = 10.0 mL of stock solution

Therefore, 10.0 ml of stock dye solution is required

c. Using m₁v₁ = m₂v

1.00 * 10⁻⁵ * v₁ = 20.0 * 2.00 * 10⁻⁶

v₁ = 4 * 10⁻⁵/1.00 * 10⁻⁵

v₁ = 4.00 mL of stock solution

Therefore, 4.00 ml of stock dye solution is required

The procedure is then followed as in steps 2 to 4.

4 0

3 years ago