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Black_prince [1.1K]
2 years ago
15

Which method is suitable for collection a acarbon dioxide.Explain​

Chemistry
1 answer:
mel-nik [20]2 years ago
8 0

Answer:

Carbon dioxide can be collected over water. Carbon dioxide is slightly soluble in water and denser than air, so another way to collect it is in a dry, upright gas jar.

Explanation:

You might be interested in
What example of mixture could be seperated using two or more techniques? Explain your answer
STatiana [176]
You can take two liquids of different densities (how much mass is in a given volume) and pour them into a funnel. An example is oil and water. When the mixture settles, the denser liquid will be at the bottom, and drips through the funnel first. This is a separation that you can just let occur naturally.
8 0
2 years ago
A gas occupies 100 mL at 150. kPa. Find its volume at 200. kPa. You must show all your work to receive credit. Be sure to identi
PtichkaEL [24]

The Boyle-Mariotte's law or Boyle's law is one of the laws of gases that <u>relates the volume (V) and pressure (P) of a certain amount of gas maintained at constant temperature</u>, as follows:

PV = k

where k is a constant.

We can relate the state of a gas at a specific pressure and volume to another state in which the same gas is at different P and V since the product of both variables is equal to a constant, according to the Boyle's law, which will be the same regardless of the state of the gas. In this way,

P₁V₁ = P₂V₂

Where P₁ and V₁ is the pressure and volume of the gas to a state 1 and P₂ and V₂ is the pressure and volume of the same gas in a state 2.

In this case, in the state 1 the gas occupies a volume V₁ = 100 mL at a pressure of P₁ = 150 kPa. Then, in the state 2 the gas occupies a volume V₂ (that we must calculate through the boyle's law) at a pressure of P₂ = 200 kPa.  Substituting these values in the previous equation and clearing V₂, we have,

P₁V₁ = P₂V₂ → V₂ = \frac{P1V1}{P2}  

→ V₂ = \frac{100 mL x 150 kPa}{200 kPa}

→ V₂ = 75 mL

Then, the volume occupied by the gas at 200 kPa is V₂ = 75 mL

6 0
3 years ago
In E. coli, the enzyme hexokinase catalyzes the reaction: Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Ke
kondor19780726 [428]

Answer:

Explanation:

Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.

In the living E. coli cells,

[ATP] = 7.9 mM;

[ADP] = 1.04 mM,

[glucose] = 2 mM,

[glucose 6-phosphate] = 1 mM.

Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.

The reaction is given as

Glucose + ATP → glucose 6-phosphate + ADP

Now reaction quotient for given equation above is

q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}

q=\frac{(1mm)\times (1.04 mm)}{(7.9mm)\times (2mm)} \\\\=6.582\times 10^{-2}

so,

q ⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable  until q = Keq

4 0
3 years ago
Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks
charle [14.2K]

Answer:

pH  = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH-   Ksp = 3X10^-15

Zn+2 + 4OH-   --> Zn(OH)4-2   Kf = 2X10^15

K = Ksp X Kf

  = 3*2*10^-15 * 10^15

  = 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

                Zn(OH)₂ + 2OH⁻(aq)  --> Zn(OH)₄²⁻(aq)

Initial:           0             0.3                      0

Change:                      -2x                     +x

Equilibrium:               0.3 - 2x                 x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²  

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

                                    = 0.3 -(2*0.089)

                                  = 0.122

pOH = -log[OH⁻]

         = -log 0.122

          = 0.91

pH = 14-0.91

     = 13.09

4 0
3 years ago
The Law of _____states that substances combine in predictable proportions and that excess reactants remain unchanged.
RideAnS [48]

Answer: definite proportions.


Explanation:


1) The definite proportions law states that compounds will always have the same kind of atoms (elements) in the same mass proportion (ratios).


2) For example, a molecule of water will alwys have the same mass ratio of hydrogen atoms to oxygen atoms. That is what permits to obtain the chemical formula of the water molecule as H₂O.


The mass of the two hydrogen atoms will be in a fixed ratio respect to the mass of the oxygen atoms.


Then, if you have one reactant in less proportion than the other, respect to the ratio stated by the chemical formula of water, the former will react completely (it is the limiting reactant) with the corresponding (proportional) mass of the later. Then there will be an excess of the later reactant which will not react (will remain unchanged).


The reactants can only react in the proportion defined by the chemical formulas of the final products.

4 0
3 years ago
Read 2 more answers
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