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4 years ago

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Calcium oxide (cao) can react with carbon dioxide (co2) in a synthesis reaction. Select the most likely products of this reactio

n. Cao + co2 → ? Caco + o2 caco3 ca + co2 + o

2 answers:

alexgriva [62]4 years ago

8 0

The answer is B.) CaCO3

Triss [41]4 years ago

4 0

Answer: Option (b) is the correct answer.

Explanation:

A synthesis reaction is the reaction in which two compounds or atoms combine together to result in the formation of a single compound.

For example, $CaO + CO_{2} \rightarrow CaCO_{3}$

Thus, here CaO and $CO_{2}$ combine together to result in the formation of calcium carbonate compound. Therefore, it is a synthesis reaction.

Therefore, we can conclude that the most likely products of Cao + co2 → CaCO3 reaction.

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Which of the following correctly describes one way that photosynthetic plants are involved in the carbon cycle? Choose 1 answer:

brilliants [131]

Answer:

The correct answer is - (Choice A) A Photosynthetic plants can move carbon from the atmosphere into an ecosystem’s food web.

Explanation:

Photosynthesis is the process performed by photosynthetic or green plants in presence of light atmospheric carbon dioxide and water converted into glucose (6 carbon compound) and oxygen.

This glucose or sugar is utilized by the plants that are producers in the ecosystem food web. These producers are fed by consumers and carbon molecules transfer from all the trophic levels in the food chain or food web in the process.

7 0

3 years ago

CaBr + KOH – Ca(OH), + KBr (balance first) What mass, in grams, of

neonofarm [45]

Answer:

129.73 g of CaBr₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CaBr₂ + 2KOH –> Ca(OH)₂ + 2KBr

Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of CaBr₂ = 40 + (80×2)

= 40 + 160

= 200 g/mol

Mass of CaBr₂ from the balanced equation = 1 × 200 = 200 g

Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)

= 40 + 2(17)

= 40 + 34

= 74 g/mol

Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

SUMMARY :

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Therefore, Xg of CaBr₂ will react to produce 48 g of Ca(OH)₂ i.e

Xg of CaBr₂ = (200 × 48)/74

Xg of CaBr₂ = 129.73 g

Thus, 129.73 g of CaBr₂ were consumed.

6 0

3 years ago

What is the expected oxidation state for the most common ion of element 2

lozanna [386]

Answer: 1+

Justification:

The ionization energies tell the amount of energy needed to release an electron and form a ion. The first ionization energy if to loose one electron and form the ion with oxidation state 1+, the second ionization energy is the energy to loose a second electron and form the ion with oxidation state 2+, the third ionization energy is the energy to loose a third electron and form the ion with oxidation state 3+.

The low first ionization energy of element 2 shows it will lose an electron relatively easily to form the ion with oxidations state 1+.

The relatively high second ionization energy (and third too) shows that it is very difficult for this atom to loose a second electron, so it will not form an ions with oxidation state 2+. Furthermore, given the relatively high second and third ionization energies, you should think that the oxidation states 2+ and 3+ for element 2 never occurs.

Therefore, the expected oxidation state for the most common ion of element 2 is 1+.

3 0

3 years ago

The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A

s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $m^3 /d$

$= 80 \times 10^5 \ I/d$

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L$

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d$

If 60 mg of alum $[Al_2(SO_4)_3.14 H_2O]$ required for one litre of the water treatment.

So Alum required for $80 \times 10^5 \ I/d$

$= 80 \times 15^5 \ I/d \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d$

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $= 60 \times 0.234 \text{ of alum ppt. per litre}$

$= 14.04$ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

= 112.32 kg/d ppt of alum

Daily total solid load is $= 144 \ kg/d + 112.32 \ kg/d$

= 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234$

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$

After addition, the aluminium hydroxide pH of water will increase due to increase in $OH^-$ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm$

And the pH is reduced into the range of 5.9 to 6.4

6 0

3 years ago

What is the concentration of h+ ions in a solution of hydrochloric acid that was prepared by diluting 35.0 ml of concentrated (1

Alekssandra [29.7K]

Hydrochloric acid ionisation is as follows;
HCl ---> H⁺ + Cl⁻
HCl is a strong base so there's complete dissociation of acid to H⁺ ions
The number of HCl moles is equivalent to number of H⁺ ions present
1 L of solution contains - 11.6 moles of H⁺ ions
In 35 ml number of moles - 11.6 mol/L / 1000 ml x 35 ml = 0.406 mol
This number of moles are dissolved in 500 ml
therefore molarity = 0.406 mol /500 ml x 1000 ml = 0.812 M

4 0

3 years ago