All categories

2 years ago

What color would a bromocresol green solution be at ph=7 ?

1 answer:

8 0

Bromocresol Green belongs to a class of dyes called as Sulfonephthaleins. This dye imparts different colours at different pH's.

Colors at different pH:
 Bromocresol Green has a transition pH range between 3.8 and 5.4.

Color Below pH 3.8:
Bromocresol Green imparts Yellow color below pH 3.8.

Color Above pH 5.4:
Bromocresol Green imparts Blue color above pH 5.4.

Due to different colors at different pH this compound is used at Indicator in quantitative analysis techniques like Titration.

Result:
Bromocresol Green will give Blue color at pH 7.

You might be interested in

A car rolls down a ramp. What is the force acting on the car that causes the movement down the ramp?

Komok [63]

The answer to your question is gravity. you are welcome. :-D

6 0

3 years ago

A 50.0 g sample of liquid water at 25.0 degree C is mixed with 29.0 g of water at 45 degree C. The final temperature of the wate

kotegsom [21]

Answer: The final temperature of water is 32.3°C

Explanation:

When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid water)

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of solution 1 (liquid water) = 50.0 g

$m_2$ = mass of solution 2 (liquid water) = 29.0 g

$T_{final}$ = final temperature = ?

$T_1$ = initial temperature of solution 1 = 25°C = [273 + 25] = 298 K

$T_2$ = initial temperature of solution 2 = 45°C = [273 + 45] = 318 K

c = specific heat of water= 4.18 J/g.K

Putting values in equation 1, we get:

$50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K$

Converting this into degree Celsius, we use the conversion factor:

$T(K)=T(^oC)+273$

$305.3=T(^oC)+273\\T(^oC)=(305.3-273)=32.3^oC$

Hence, the final temperature of water is 32.3°C

7 0

3 years ago

A Lewis acid is a substance that can

katrin2010 [14]

Accept a pair of nonbonding electrons,a Lewis acid is an electron-pair acceptor. A Lewis base is any substance, such as the OH- ion, that can donate a pair of nonbonding electrons. A Lewis base is therefore an electron-pair donor.

6 0

2 years ago

A sample of neon has a volume of 40.81 m3 at 23.5C. At what temperature, in Kelvins, would the gas occupy 50.00 cubic meters? As

mezya [45]

At $\fbox{\begin \\363 K \end{minispace}}$ temperature, a sample of neon gas will occupy $50.00 \text{ m}^{3}$ volume.

Further Explanation:

The given problem is based on the concept of Charles’ law. Charles’ law states that “at constant pressure and fixed mass the volume occupied an ideal gas is directly proportional to the Kelvin temperature.”

Mathematically the law can be expressed as,

$\fbox{ \begin \\ V \propto T \end{minispace}}$

Or,

$\frac{V}{T}=k$

Here, V is the volume of the gas, T is Kelvin temperature, and k is proportionality constant.

Given information:

The initial volume of neon gas is $40.81 \text{ m}^{3}$ .

The final volume of neon gas is $50.00 \text{ m}^{3}$ .

The initial temperature value is $23.5 \text{ } ^{\circ} \text{C}$ .

To calculate:

The final temperature

Given Condition:

The pressure is constant.

Mass of gas is fixed.

Solution:

Step 1: Modify the mathematical expression for Charles’ law for two different temperature and volume values as follows:

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

Here,

$V_{1}$ is the initial volume of the gas.

$V_{2}$ is the final volume of the gas.

$T_{1}$ is the initial temperature of the gas.

$T_{2}$ is the final temperature of the gas.

Step 2: Rearrange equation (2) for .

$\fbox {\begin \\T_{2}=\frac{(V_{2}) \times (T_{1})}{V_{1}}\\\end{minispace}}$ …… (2)

Step 3: Convert the given temperature from degree Celsius to Kelvin.

The conversion factor to convert degree Celsius to Kelvin is,

$T(\text{K}) = T(^{\circ}\text{C}) + 273.15$ …… (3)

Substitute $23.5\text{ }^{\circ} \text{C}$ for $T(^{\circ}\text{C})$ in equation (3) to convert temperature from degree Celsius to Kelvin.

$T(\text{K}) = 23.5 \text{ } ^{\circ} \text{C} + 273.15\\T(\text{K})= 296.65 \text{ K}$

Step 4: Substitute $40.81 \text{ m}^{3}$ for $V_{1}$ , $50.00 \text{ m}^{3}$ for $V_{2}$ and $296.65 \text{ K}$ for $T_{1}$ in equation (2) and calculate the value of $T_{2}$ .

$T_{2}=\frac{(50.00 \text{ m}^{3}) \times (296.65 \text{ K})}{40.81 \text{ m}^{3}}\\T_{2}=363.45 \text{ K}\\T_{2} \approx 363 \text{ K}$

Important note:

The temperature must be in Kelvin.

The condition of fixed mass and fixed pressure must be fulfilled in order to apply Charles’ law.

Learn More:

1. Gas laws brainly.com/question/1403211

2. Application of Charles’ law brainly.com/question/7434588

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: States of matter

Keywords: neon, volume, occupies, temperature, Kelvin, degree Celsius, Charle’s law, constant pressure, fixed mass, 40.81 m^3 , 50.00 m^3 , 23.5 degree C , celsius , 363 K , sates of matter, initial volume, final volume, initial temperature, final temperature, V1 , V2 , T1 , T2 .

5 0

2 years ago

Read 2 more answers

Adi loves the smell of lavender, so she decides to press a few petals of the flower until they release a few drops of oil. She w

soldier1979 [14.2K]

Answer:

See Explanation

Explanation:

What Adi failed to realize is that the oily substance that was obtained from lavender consists of a mixture of substances. It is not only the required fragrance that is present in the extract.

This experiment will not work because those other components in the mixture may be erroneously identified when they show up in the mass spectrum of the extract and may be mistaken for the fragrance in question.

Hence the experiment will not work because; if some kind of separation method is not used to identify other impurities in the oil, many other substances may be mistaken for the actual fragrance.

4 0

2 years ago