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UkoKoshka [18]
3 years ago
11

Multiple Choice Which of the following numbers are equivalent to 20%? Choose all that apply. A. 0.002 B. The fraction is 1 over

4. C. 0.20 D. one-fifth
Chemistry
1 answer:
Mars2501 [29]3 years ago
6 0

Answer:

the answer is d. 1/5 because if you times the bottom number by 20 you will get 100 and the top number wil be a 20 which is 20%

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PLEASE HURRY
Evgen [1.6K]

<em>Answer:</em>

4) the one that is reduced, which is the oxidizing agent

<em>Explanation:</em>

<em>An oxidizing agent is one that causes oxidation by gaining electrons from another atom/molecule. </em>

6 0
3 years ago
Adding heat to an exothermic reaction will have what effect? Reactants ↔ products + heat View Available Hint(s) Adding heat to a
Alina [70]

Answer:

Causes the equilibrium to shift to the left, in favor of making more reactants, and K decreases.

Explanation:

Le Châtelier's principle states that if there is a stress in equilibrium, the reaction will shift to restore the equilibrium. An exothermic reaction loses heat for the surroundings, so the equilibrium must be represented as:

Reactants ⇔ Products + Heat

Then, when more heat is added, to restore the equilibrium, the reaction shift to the left ("consuming" heat), in favor of making more reactants.

The equilibrium constant (K) is:

K = [Products]/[Reactants]

So, [Reactants] will increase, and K must decrease.

3 0
3 years ago
Please somebody help!!!!!!!!!!!!!
natta225 [31]

Answer:

1.Materials

2. Hypothesis

3.Question

4.Procedure

5.Further questions

6.Results and Data

7.Conclusion

Explanation:

4 0
3 years ago
At normal blood pH pH (7.4), hemoglobin is 80 80 % saturated at a partial pressure of oxygen ( O 2 O2 ) of 40 mmHg 40 mmHg . Use
schepotkina [342]

Answer:

An example of oxygen–hemoglobin (O2–Hb) dissociation curves from (A) one penguin at pH 7.5, 7.4 and 7.3, and (B) the emperor penguin, the bar-headed goose (Anser indicus) (Black and Tenney, 1980) and the domestic duck (Anas platyrhynchos, forma domestica) (Hudson and Jones, 1986) at pH 7.4. Note that as for the bar-headed goose, the O2–Hb dissociation curve of the emperor penguin is significantly left-shifted as compared with the domestic duck (and most birds). The bar-headed goose photo is courtesy of Graham Scott; the domestic duck photo is by Maren Winter (licensed under the terms of the GNU Free Documentation License, Version 1.2 or any later version); the penguin photo is by J.M.

Explanation:

The resulting regression equations from the plots of log[SO2/(100–SO2)] vs log(PO2) (all saturation points, all penguins combined) were:

pH 7.5: log[SO2/(100–SO2)] = 2.92589 × log(PO2) – 4.24338 (N=43, r2=0.98, P<0.0001),

pH 7.4: log[SO2/(100–SO2)] = 2.94767 × log(PO2) – 4.39858 (N=70, r2=0.98, P<0.0001),

pH 7.3: log[SO2/(100–SO2)] = 3.04945 × log(PO2) – 4.72019 (N=38, r2=0.99, P<0.0001),

pH 7.2: log[SO2/(100–SO2)] = 3.15958 × log(PO2) – 4.97618 (N=9, r2=0.99, P<0.0001).

5 0
3 years ago
Consider the reaction: 3Co2+(aq) + 6NO3¯(aq) + 6Na+(aq) + 2PO43¯(aq) â Co3(PO4)2(s) + 6Na+(aq) + 6NO3¯(aq) Identify the net i
irina [24]

Answer:

D. 3 Co²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s)

Explanation:

The complete ionic equation includes all the ions and the insoluble species. Let's consider the following complete ionic equation.

3 Co²⁺(aq) + 6 NO₃⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s) + 6 Na⁺(aq) + 6 NO₃⁻(aq)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the insoluble species. The corresponding net ionic equation is:

3 Co²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s)

3 0
3 years ago
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