The overall long-term effects of air pollution are not yet certain. True False

A sample of gas starts at 1.00 atm, 0.00 degrees Celsius, and 30.0 mL. What is the

lara31 [8.8K]

Answer:

pV= nRT

Explanation:

(p1 × V1)/ T1/ (p2 × V2)/ T2

4 0

3 years ago

Draw a structural formula for the alkene you would use to prepare the alcohol shown by hydroboration/oxidation.

Nastasia [14]

Answer:

See explanation

Explanation:

The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.

This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).

In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.

Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.

Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.

5 0

3 years ago

An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction At

anygoal [31]

Answer: [N2]₀ = 10M and [H2]₀ = 11M

Explanation: To calculate the initial concentration, you would have to set up an ICE table, which is an organized way of tracking known quantities or the ones you want to find. ICE stands for:

I is initial amount;

C is change in concentration;

E is for equilibrium concentration;

For the mixture,

N2 3H2 2NH3

I [N2]₀ [H2]₀ 0

C - x -3x +2x

E [N2]₀ - x =8 [H2]₀ - 3x =5 2x =4

With the product, we can find "x":

2x=4

x=2M

With x=2, find the concentrations:

[N2]₀ - x = 8

[N2]₀ = 10M

[H2]₀ - 3x = 5

[H2]₀ = 11M

The initial concentrations of nitrogen gas [N2] is 10.0 M and of hydrogen gas [H2] is 11.0 M.

8 0

3 years ago

Read 2 more answers

an aqueous solution contains 1.2mM of total ions. if the solution is NaCl (aq), what is the concentration of chloride ions?

Genrish500 [490]

Explanation:

As the total concentration is given as 1.2 mM. And, it is also given that salt present in the solution is NaCl.

As sodium chloride is an ionic compound so, when it is added to water then it will dissociate into sodium and chlorine ions as follows.

$NaCl \rightarrow Na^{+} + Cl^{-}$

So, it means in total there will be formation of 2 ions when one molecules of NaCl dissociates.

Therefore, concentration of chlorine ions will be calculated as follows.

Concentration of $Cl^{-}$ ions = $\frac{1.2mM}{2}$

= 0.6 mM

Thus, we can conclude that the concentration of chloride ions is 0.6 mM.

5 0

3 years ago

Which one of the following statements is not true concerning 2.00 L of a 0.100 M solution of Ca3(PO4)2?

Len [333]

Answer: The correct answer is Option B.

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .......(1)

For A:

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 2.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{2.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 2.00L)=0.200mol$

Moles of calcium phosphate = 0.200 moles

For B:

1 mole of calcium phosphate contains 3 moles of calcium atoms, 2 moles of phosphate atoms and 8 moles of oxygen atoms.

So, 0.200 moles of calcium phosphate will contain = $(8\times 0.200)=1.6$ moles of oxygen atoms.

Moles of oxygen atoms = 1.6 moles

For C:

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 1.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{1.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 1.00L)=0.100mol$

Moles of calcium ions = $(0.100\times 3)=0.300$ moles

For D:

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 5.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{5.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 5.00L)=0.500mol$

Moles of phosphorus atoms = $(0.500\times 2)=1.00$ moles

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of atoms

Number of phosphorus atoms in 0.500 moles of calcium phosphate will be = $(1.00\times 6.022\times 10^{23})=6.022\times 10^{23}$

For E:

1 mole of calcium phosphate contains 3 moles of calcium ions and 2 moles of phosphate ions.

So, 0.200 moles of calcium phosphate will contain = $(3\times 0.200)=0.600$ moles of calcium ions

Moles of calcium ions = 0.600 moles

Hence, the correct answer is Option B.

6 0

3 years ago