Please help with this chemistry because I’m confused

A student did not read the directions to the experiment properly and mixed up where to place the NaOH solution and the vinegar.

Keith_Richards [23]

Answer:

The answer is "Only at the end, a transformation of rose to color is made ".

Explanation:

In this student puts its vinegar in the titrator, and NaOH throughout the beaker, which implies phenolphthalein has also been poured into water.

Phenolphthalein does have a pH range of 8.3-10 (approx). It's indeed pink in the basic medium therefore, the formulation becomes pink throughout the color by adding phenolphthalein to NaOH.

It is beginning of vinegar was its beginning of neutralization of NaOH from the titrator through full neutralization, a single piece of vinegar is added to a solution as well as the rose solution is lost throughout the beaker.

7 0

3 years ago

How is the temporory hardness of water remove bye boiling mathod

IgorLugansk [536]

Answer:

The temporary hardness of water can be removed by boiling. The bicarbonates get converted to insoluble carbonates and settle down at the bottom.

Calcium bicarbonate -------> Calcium carbonate [insoluble] + Water + Carbon dioxide.

Ca[HCO3]2 -----> Ca CO3 + H2O + CO2

5 0

3 years ago

A bicycle pump contains 250 mL of air at a pressure of 102 kPa. If the pump is

CaHeK987 [17]

Considering the Boyle's law, the new volume becomes 1,148,344.444 mL or 1148.34 L.

<h3>Boyle's law</h3>

Boyle's law relates pressure and volume, stating that the volume occupied by a given mass of gas at constant temperature is inversely proportional to pressure. This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

P×V=k

Now it is possible to assume that you have a certain volume of gas V1 which is at a pressure P1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the pressure will change to P2, and the following is true:

P1×V1=P2×V2

<h3>New volume</h3>

In this case, you know:

P1= 102 kPa= 10335.1 atm (being 1 kPa= 101.325 atm)
V1= 250 mL
P2= 2.25 atm
V2= ?

Replacing in the Boyle's law:

10335.1 atm ×250 mL= 2.25 atm×V2

Solving:

(10335.1 atm ×250 mL)÷ 2.25 atm= V2

Finally, the new volume becomes 1,148,344.444 mL or 1148.34 L.

Learn more about Boyle's law:

brainly.com/question/4147359

#SPJ1

3 0

2 years ago

Consider the interaction of two hydrogen ls atomic orbitals of the same phase. Which statements below is an incorrest descriptio

Andreyy89

Answer:

B) The molecular orbital formed is lower in energy than a hydrogen 1s atomic orbital.

Explanation:

When two atoms of hydrogen come close to each other , there is formation of molecular orbital . Due to overlap of 1 s orbital of one and 1 s orbital of another atom , two molecular orbitals are formed . One of these molecular orbital has energy less than 1 s atomic orbital . It is called 1 s sigma bonding molecular orbital . The other molecular orbital has energy more than 1 s atomic orbital . It is called antibonding molecular orbital . Two electrons occupy bonding sigma molecular orbital .

So , the statement that "the molecular orbital formed is lower in energy than a hydrogen 1s atomic orbital " is wrong .

4 0

3 years ago

Sabendo que os calores de combustão do enxofre monoclínico e do enxofre rômbico são, respectivamente, - 297,2 kJ/mol e - 296,8 k

liq [111]

Responda:

+ 0,9kJ / mol

Explicação:

Dados os calores de combustão do enxofre monoclínico e enxofre rômbico como - 297,2 kJ / mol e - 296,8 kJ / mol, respectivamente para a variação na transformação de 1 mol de enxofre rômbico em enxofre monoclínico conforme mostrado pela equação;

S (mon.) + O2 (g) -> SO2 (g)

Uma vez que são todos 1 mol cada, a mudança na entalpia será expressa como ∆H = ∆H2-∆H1

Dado ∆H2 = -296,8kJ / mol

∆H1 = -297,2kJ / mol

∆H = -296,8 - (- 297,2)

∆H = -296,8 + 297,2

∆H = 297,2-296,8

∆H = + 0,9kJ / mol

Portanto, a mudança na entalpia da equação é + 0,9kJ / mol

4 0

3 years ago