Substances in a mixture keep their own properties. True or False?

When would a life scientist study a nonliving thing such as a rock or lake

VladimirAG [237]

He would study it so he could see how it interacted with the living things around it.

3 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

How to balance lead and silver acetate yields lead (||) acetate and silver

arsen [322]

The given elements put into an equation using their symbols are as follows:
Pb + $AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + Ag

Since there are 2 Pb on the right side of the equation, you would change the coefficient of Pb on the left side to 2:
2Pb + $AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + Ag

Since there are 2 Acetate on the right side of the equation, you would change the coefficient of Silver Acetate on the left side to 2:
2Pb + $2AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + Ag

Now there are 2 Silver on the left side, so you change the coefficient of Silver on the right side to 2:
2Pb + $2AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + 2Ag

That is your final equation

The coefficients are 2 + 2 = 1 + 2

3 0

3 years ago

Please help me with these questions

zaharov [31]

I’m just answering this so i can ask more questions and it has to be 20 words long so i hope you figure your problems out and merry christmas happy new year

5 0

3 years ago

About 55L of a gas in a flexible container is under a pressure of 3.2 atm and at a temperature of 520K. What is the new volume o

vampirchik [111]

Answer:

30.62 L

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 55 L

Initial pressure (P₁) = 3.2 atm

Initial temperature (T₁) = 520 K

Final temperature (T₂) = 760 K

Final pressure (P₂) = 8.4 atm

Final volume (V₂) =?

The final volume of the gas can be obtained as follow:

P₁V₁ / T₁ = P₂V₂ / T₂

3.2 × 55 / 520 = 8.4 × V₂ / 760

176 / 520 = 8.4 × V₂ / 760

Cross multiply

520 × 8.4 × V₂ = 176 × 760

4368 × V₂ = 133760

Divide both side by 4368

V₂ = 133760 / 4368

V₂ = 30.62 L

Therefore, the new volume of the gas is 30.62 L

7 0

3 years ago