PLEASE I NEED HELP ASAP

Increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose mol

Sonja [21]

Answer: The given statement is true.

Explanation:

When we increase the amount of solvent which is water in this case then it means there will occur an increase in the molecules. Hence, there will be more number of collisions to take place with increase in number of molecules.

Therefore, more is the amount of interaction taking place between the molecules of a solution more will be its rate of hydrolysis.

Thus, we can conclude that the statement increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis, is true.

8 0

3 years ago

Given the Henry’s law constant for O2(4.34*109Pa) at 25° C, calculate the molar concentration of oxygen in air-saturated and O2s

Flauer [41]

This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

Answer : The molar concentration of oxygen is, $2.67\times 10^2mol/L$

Explanation :

As we know that,

$C_{O_2}=k_H\times p_{O_2}$

where,

$C_{O_2}$ = molar solubility of $O_2$ = ?

$p_{O_2}$ = partial pressure of $O_2$ = 0.2 atm = 1.97×10⁻⁶ Pa

$k_H$ = Henry's law constant = 4.34 × 10⁹ g/L.Pa

Now put all the given values in the above formula, we get:

$C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)$

$C_{O_2}=8.55\times 10^3g/L$

Now we have to molar concentration of oxygen.

Molar concentration of oxygen = $\frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L$

Therefore, the molar concentration of oxygen is, $2.67\times 10^2mol/L$

8 0

3 years ago

PLEASE HELPPPP IM GIVING THE TEST RN

Tom [10]

Answer:

answer is c

Explanation:

single replacement

6 0

3 years ago

When solving a problem it is important to identify your given and needed units, but it is also important to understand the relat

Angelina_Jolie [31]

Answer:

The question has some details missing. here are the details ; Given the following ;

1. 43.2 g of tablet with 20 cm3 of space

2. 5 cm3 of tablets weighs 10.8 g

3. 5 g of balsa wood with density 0.16 g/cm3

4. 150 g of iron. With density 79g/cm 3

5. 32 cm3 sample of gold with density 19.3 g/cm3

6. 18 ml of cooking oil with density 0.92 g/ml

Explanation:

Appropriate for calculating mass

32 cm3 sample of gold with density 19.3 g/cm3

18 ml of cooking oil with density 0.92 g/ml

Appropriate for calculating volume

5 g of balsa wood with density 0.16 g/cm3

150 g of iron. With density 79g/cm 3

Appropriate for calculating density

43.2 g of tablet with 20 cm3 of space

5 cm3 of tablets weighs 10.8 g

3 0

3 years ago

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide dissolved in molten cryolit

natka813 [3]

The given question incomplete, the complete question is:

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,03) dissolved in molten cryolite (Na, Alts).re in the reduction of the Al, o, to pure aluminum. Suppose a current of 1800. A is passed through a Hall-Heroult cell for 37.0 seconds. Calculate the mass of pure aluminum produced Be sure your answer has a unit symbol and the correct number of significant digits.

Answer:

The correct answer is 6.2114 grams.

Explanation:

Based on the given question, the value of current or I have given is 1800 amperes, the time given is 37 seconds, and there is a need to find the mass of the pure aluminum generated in the process. Mass or weight can be determined by using Faraday's first law equation, that is, w = MIt/nF.

Here, M is the atomic mass, w is the weight of the substance deposited, t is time, I is current, n is the number of moles of the electron, and F is the Faraday's constant, which is 96500 C. In the process mentioned in the question, aluminum oxide is reduced to give rise to pure aluminum, and in the process 3 electrons are gained. So, the value of n will be 3. The M or the atomic mass of Al is 27 gm per mole. Now putting the values in the equation we get,

w = 27*1800*37 / 3*96500

w = 1798200 / 289500

w = 6.2114 grams

Hence, pure aluminum produced in the process is 6.2114 grams.

7 0

3 years ago