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telo118 [61]
3 years ago
11

En las siguientes reacciones identifique cuales sustancias corresponden a ácidos y bases de Arrhenius y cuales a ácidos y bases

de Brønsted – Lowry
a. HBr → H+ + Br-
b. NH4 + OH H2O+ + NH3
c. HCO3+ H2O CO3- + H3O+
d. KOH → K+ + OH-
e. HNO3 → H+ + NO3-
f. Al (OH)3 → Al+ + 3OH-
g. H2SO4- + H2O H3O+ + HSO4
Chemistry
1 answer:
ser-zykov [4K]3 years ago
7 0

Answer:

do you have an english version

Explanation:

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Answer: A. There will be fewer producers in that area.

Explanation: Answer A is correct because all plants, including grass, is a producer. Since there is an increase in grass eating animals (Antelope), more grass will be consumed, decreasing the amount of grass aka producer.

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2 years ago
What is infrared energy? simplify it
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Infrared energy is the energy of light between microwave radiation and Ultraviolet radiation
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Aluminum metal reacts with dilute sulfuric acid to produce aluminum sulfate and hydrogen gas. what mass of aluminum will react w
SOVA2 [1]
Aluminium reacts with dilute sulfuric acid based on the following reaction:
<span>2Al + 3H2SO4 ..............> Al2 (SO4)3 + 3H2

From the periodic table:
mass of aluminium = 27 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
mass of sulfur = 32 grams

Therefore:
molar mass of aluminium = 27 grams
molar mass of sulfuric acid = 2(1) + 32 + 4(16) = 98 grams

From the balanced chemical equation:
2 moles of aluminium react with 3 moles of dilute sulfuric acid.
This means that 34 grams of Al react with 294 grams of the acid

To get the amount  of aluminium that reacts with </span><span>5.890 g of sulfuric acid, we will do cross multiplication as follows:
</span>amount of Al = (<span>5.890 x 34) / 294 = 0.6811 grams</span>
3 0
3 years ago
A chemist requires 0.802 mol Na2CO3 for a reaction. How many grams does this correspond to?
Komok [63]

Answer:

Ok:

Explanation:

So grams = mols*MolarMass. Here, MolarMass (MM) = 105.99g which can be found using the periodic table. mols is given to be 0.802. We can then plug in to get that it corresponds to 85.0g.

7 0
3 years ago
Calculate the molarity of 13.1 g of NaCl in 727 mL of solution
valentina_108 [34]

Answer:

The molarity of the solution is 0,31 M

Explanation:

We calculate the weight of 1 mol of NaCl from the atomic weights of each element of the periodic table. Then, we calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case NaCl) in 1000ml of solution (1 liter)

Weight 1 mol NaCl= Weight Na + Weight Cl= 23 g + 35, 5 g= 58, 5 g

58, 5 g-----1 mol NaCl

13,1 g ---------x= (13,1 g x 1 mol NaCl)/58, 5 g= 0, 224 mol NaCl

727 ml solution------ 0, 224 mol NaCl

1000ml solution------x= (1000ml solutionx0, 224 mol NaCl)/727 ml solution

x=0,308 mol NaCl---> <em>The solution is 0,31 molar (0,31 M)</em>

5 0
3 years ago
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