Which type of chemical bond requires the least amount of energy to break?

Which of the following describe chemical properties?

stealth61 [152]

<span>chemical properties is the changes in chemical manner of a
substances. Antacid tablets neutralize stomach acid is a chemical properties (it is an acid) which produce new products... salt and water
The others are physical properties</span>

3 0

3 years ago

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Using the mmoles of (35)-2,2,-dibromo-3,4-dimethylpentane calculated earlier and the molecular weight of the product (962 g/mol)

Sedbober [7]

Answer:

The yield of the product in gram is $\mathsf{{w_P}=0.26 \ gram}$

Explanation:

Given that:

the molecular mass weight of the product = 96.2 g/mol

the molecular mass of the reagent (3S)-2,2,-dibromo-3,4-dimethylpentane is 257.997 g

given that the millimoles of the reagent = 2,7 millimoles = $2.7 \times 10^{-3} \ moles$

We know that:

Number of moles = mass/molar mass

Then:

$2.7 \times 10^{-3} = \dfrac{ mass}{257.997}$

$mass = 2.7 \times 10^{-3} \times 257.997$

mass = 0.697

Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100

i.e

Theoretical yield = $\dfrac{n_P}{n_R}\times 100\%$

where;

$n_P = \dfrac{w_P}{m_P}$ and $n_R = \dfrac{w_R}{m_R}$

Theoretical yield = $\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%$

Given that the theoretical yield = 100%

Then:

$100\% =\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%$

$\dfrac{w_P}{m_P}=\dfrac{w_R}{m_R}$

${w_P}=\dfrac{w_R \times m_P}{m_R}$

where,

$w_P$ = derived weight of the product

$m_P =$ the molecular mass of the derived product

$m_R =$ the molecular mass of the reagent

$w_R$ = weight in a gram of the reagent

${w_P}=\dfrac{w_R \times m_P}{m_R}$

${w_P}=\dfrac{0.697 \times 96.2}{257.997}$

$\mathsf{{w_P}=0.26 \ gram}$

8 0

2 years ago

How old are the lunar highlands? 1. 1.8 to 2.6 billion years old 2. 3.1 to 3.8 billion years old 3. 4.0 to 4.3 billion years old

Anastaziya [24]

They are about 4.5 billion years old. Hope this helps.

4 0

3 years ago

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Assuming all the nh3 dissolves and that the volume of the solution remains at 0.300 l , calculate the ph of the resulting soluti

barxatty [35]

<span>Assume p=735 Torr V= 7.6L R=62.4 T= 295 PV-nRT (735 Torr)(7.60L)= n (62.4Torr-Litres/mole-K)(295K) 0.30346 moles of NH3 Find moles 0.300L solution of 0.300 M HCL = 0.120 moles of HCL 0.30346 moles of NH3 reacts with 0.120 moles of HCL producing 0.120 moles of NH4+ ION, and leaving 0.18346 mole sof NH3 behind Find molarity 0.120 moles of NH4+/0.300L = 0.400 M NH4+ 0.18346 moles of NH3/0.300L = 0.6115 M NH3 NH4OH --> NH4 & OH- Kb = [NH4+][OH]/[NH4OH] 1.8 e-5=[0.300][OH-]/[0.6115] [OH-]=1.6e-5 pOH= 4.79 PH=9.21 .</span>

3 0

3 years ago

The mass of cupric sulfate needed to make .5 liters of a .5M solution is

dalvyx [7]

Answer:

C

Explanation:

3 0

3 years ago