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2 years ago

What happens to the water and oil mixture when air is introduced

1 answer:

4 0

well, the oil will come up on the surface of the water and lay there but introducing air does not do much. hope that helps

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What is the oh- in a solution with a poh of 5.71

Rudik [331]

Answer:- The hydroxide ion concentration of the solution is $1.95*10^-^6$ .

Solution:- The formula used to calculate pOH from hydroxide ion is:

$pOH=-log[OH^-]$

When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:

$[OH^-]=10^-^p^O^H$

pOH is given as 5.71 and we are asked to calculate hydrogen ion concentration. Let's plug in the given value in the formula:

$[OH^-]=10^-^5^.^7^1$

$[OH^-]$ = 0.00000195 or $1.95*10^-^6$

So, the hydroxide ion concentration of the solution is $1.95*10^-^6$ .

3 0

3 years ago

I need help with homework

stellarik [79]

The equation that conforms with the particle model is 4C + 6O2 ----> 4CO2 + 2O2.

<h3>What is a chemical reaction?</h3>

The term chemical reaction has to do with the combination of substances to give products. We know that reactants and products must follow the law of conservation of mass.

Hence, the equation that conforms with the particle model is 4C + 6O2 ----> 4CO2 + 2O2.

Learn more about chemical reaction:brainly.com/question/22817140

#SPJ1

4 0

1 year ago

Which term describes the amount of order in a system

Ivahew [28]

Answer:

entropy

Explanation:

3 0

3 years ago

Most chemical bonds are

Lilit [14]

Answer:

partly ionic and partly covalent

3 0

2 years ago

A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system

AleksAgata [21]

To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

$-m\cdot c_e\cdot \Delta T = m_w\cdot c_e_w\cdot \Delta T_w$

Clearing the specific heat of the unknown substance:

$c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}$

$c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}$

6 0

2 years ago