2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at 760.0 mmHg

what is the approximate ph at the equivalence point of a weak acid-strong base titration if 25 ml of aqueous formic acid require

vivado [14]

Answer:

pH at equivalence point is 8.52

Explanation:

$HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O$

1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of $HCOO^{-}$

So, moles of NaOH used to reach equivalence point equal to number of moles $HCOO^{-}$ produced at equivalence point.

As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.

So, moles of $HCOO^{-}$ produced = $\frac{29.80\times 0.3567}{1000}moles=0.01063moles$

Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL

So, at equivalence point concentration of $HCOO^{-}$ = $\frac{0.01063\times 1000}{54.80}M=0.1940M$

At equivalence point, pH depends upon hydrolysis of $HCOO^{-}$ . So, we have to construct an ICE table.

$HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}$

I: 0.1940 0 0

C: -x +x +x

E: 0.1940-x x x

So, $\frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}$

species inside third bracket represent equilibrium concentrations

So, $\frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}$

or, $x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0$

So, $x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}$

So, $x=3.285\times 10^{-6}M$

So, $pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52$

5 0

3 years ago

n the laboratory, two forms of sodium phosphate will be available (the monobasic monohydrate NaH2PO4·H2O, F.W. = 137.99 g/mol, a

const2013 [10]

Answer:

The compound you will use is the Dibasic phosphate

Explanation:

Simple phosphate buffer is used ubiquitously in biological experiments, as it can be adapted to a variety of pH levels, including isotonic. This wide range is due to phosphoric acid having 3 dissociation constants, (known in chemistry as a triprotic acid) allowing for formulation of buffers near each of the pH levels of 2.15, 6.86, or 12.32. Phosphate buffer is highly water soluble and has a high buffering capacity,

In this case the most efficient way is to disolve the dibasic compound which in the reaction with the water will form the monobasic phosphate.

To make the buffer you have to prepare the amount of distillate water needed, disolve the dibasic phospate, and then adjust with HCl or NaOH depending on the pH needed.

6 0

3 years ago

(CH3)2-CH-CH2-O(CH3)3IUPAC NAME

bazaltina [42]

Answer:

1-(tert-butoxy)-2-methylpropane

Note: there is a mistake in formula, the correct formula is (CH₃)₂-CH-CH₂-O-C(CH₃)₃ not (CH₃)₂-CH-CH₂-O(CH₃)₃, because oxygen is a divalent compound.

Explanation:

Structural formula is attached

IUPAC naming rules

1. start numbering the chain from the functional group. In this compound we start from oxygen side.

2. Here we can see that at position 1 there is an oxy group along with a tertiary carbon having three methyl groups. So we write it as 1-tert-butoxy. Which means that there is a methoxy group at position 1 along with a tertiary carbon.

3. At position 2 we can see that there is a methyl group attached to the main chain, so we write it as 2-methyl.

4. Now we count the total number of carbons in the main chain. As we can see that there are 3 carbons in the remaining or parent chain, so we write it as propane

5. So the IUPAC name of the compound will be 1-(tert-butoxy)-2-methylpropane

3 0

3 years ago

What is the volume, in liters, of 0.500 mol of c3h3 gas at stp?

jarptica [38.1K]

At STP conditions the volume of 1 mol of any ideal gas will be 22.4L

0.500 mol C3H3 x 22.4L / 1 mol = 11.2 L

8 0

3 years ago

The process of ____ occurs when magma, lava, or a solute in a solution becomes a solid and the internal components will arrange

satela [25.4K]

Answer:

D. Crystallization

Explanation:

Let's clarify the irrelevant terms first.

unification: This term has nothing to do with chemistry at all
lithification: When the problem mentions magma and lava, you might think that this term is related to the process here. However, 'lithification' do have a precise meaning in geology. It refers to the process where sediments collapses into one single rock under pressure, which has nothing to do with the process mentioned here.

Now, for 2 terms that might confuse you: 'solidification' and 'crystallization' these also has precise scientific definition

Solidification is defined the process where substances in liquid phase changes its phase to solid. On first glance, this answer might seems correct, and yes, it is correct for this question. But not the most correct.

The keyword here is

'the internal components will arrange its self in an organized pattern.'

Crystallization is a special case of Solidification where the atoms or molecules of liquid solidify by spontaneously arrange themselves in periodic, ordered, and organized pattern. It might or might not happen during solidification depending on cooling rate, viscosity of liquid, and other factors.

So, Crystallization is the most correct answer here.

4 0

3 years ago