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3 years ago

Which compounds are lewis bases? select all that apply. h―c≡c―h h− nh3 ch3ch2ch3?

1 answer:

8 0

The Lewis bases are those which can donate electron pair to an acceptor. These electrons can be a pair of electron that is a lone pair, a negative charge or electrons in a $\pi$ -bond.

In $H-C\equiv C-H$ , the presence of $\pi$ -bond make it a Lewis base.
In $H^{-}$ , the presence of negative charge makes it a Lewis base.
In $NH_3$ , the presence of lone pair on nitrogen makes it a Lewis base.
In $CH_3CH_2CH_3$ , there is no lone pair, no $\pi$ -bond, no negative charge so it is not a Lewis base.

Hence, $H-C\equiv C-H$ , $H^{-}$ , and $NH_3$ are Lewis bases.

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Select the correct answer. if two half-lives have passed since a scientist collected a 1.00-gram sample of u-235, how much u-235

Ugo [173]

Answer:

0.25 g of U-235 isotope will left .

Formula used :

where,

N = amount of U-235 left after n-half lives = ?

= Initial amount of the U-235 = 1.00 g

n = number of half lives passed = 2

0.25 g of U-235 isotope will left .

3 0

2 years ago

Is this equation balanced and in the lowest form? 4NH3 → 2N2 + 6H2

solong [7]

Answer:

B.) No, because the coefficients could be reduced to 2,1, and 3.

Explanation:

The equation is not in its lowest molar ratio form. In this case, all of the coefficients can be divided by 2 and still result in whole numbers.

As such, the correct balanced equation is:

2 NH₃ ----> N₂ + 3 H₂

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2 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro

MAVERICK [17]

Answer: The percent yield of water is 46.9 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For sulfuric acid:

Given mass of sulfuric acid = 72.6 g (Assuming)

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfuric acid}=\frac{72.6g}{98g/mol}=0.741mol$

For NaOH:

Given mass of NaOH = 77 g (Assuming)

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaOH}=\frac{77g}{40g/mol}=1.925mol$

The chemical equation for the reaction of sulfuric acid and sodium hydroxide follows:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of NaOH

0.741 moles of sulfuric acid will react with = $\frac{2}{1}\times 0.741=1.482mol$ of NaOH

As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent.

Thus, sulfuric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of water

0.741 moles of sulfuric acid will react with = $\frac{2}{1}\times 0.741=1.482mol$ of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.482 moles

Putting values in equation 1, we get:

$1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol\times 18g/mol)=26.67g$

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 12.5 g (Assuming)

Theoretical yield of water = 26.67 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{12.5g}{26.67g}\times 100\\\\\% \text{yield of water}=46.9\%$

Hence, the percent yield of water is 46.9 %

3 0

3 years ago

Indicate the type of solute-solvent interaction that should be most important in each of the following solutions.a. LiCl in wate

topjm [15]

Explanation:

a. LiCl is an ionic molecule whereas water is a polar molecule with net dipole moment in it. There LiCl in water would have an ion-dipole force of interaction.

b. Both NF3 and CH3CN have dipole moment in them, since both are polar molecule. Hence, there would be dipole-dipole interaction.

c. Here both CCl4 and benzene are non polar molecules therefore, they have London dispersion force of interaction.

d. In methylamine and water both have hydrogen bonding in them. The nitrogen of CH3NH2 forms hydrogen bond with water.

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3 years ago

During the phosphatase experiment you will use several different ph buffers. the ph value depends upon combing various ratios of

Elan Coil [88]

First calculate mass of sodium hydrogen phosphate (Na₂HPO₄) needed for solution:
n(Na₂HPO₄) = V(Na₂HPO₄) · c(Na₂HPO₄)
n(Na₂HPO₄) = 0,3 dm³ · 0,2 mol/dm³
n(Na₂HPO₄) = 0,06 mol
m(Na₂HPO₄) = 0,06 mol · 142 g/mol
m(Na₂HPO₄) = 8,52 g.
Measure that mass on scale using spoon and beaker. Than dissolve salt in beaker with distilled water and pour that solution in volumetric flask (size 300 mL), pour distilled water to graduation mark.

7 0

3 years ago