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3 years ago

What elemental classification does oxygen belong to?

2 answers:

6 0

The elemental classification for oxygen would be non-metal. It is located in period number 2 and group number 16. It has an atomic number of 8. At room temperature, it exists as a gas. Hope this answers the question. Have a nice day.

Komok [63]3 years ago

3 0

Answer: Oxygen is a non-metal.

Explanation:

Elements are distributed into 7 period and 18 Groups in a periodic table.

Elements are classified into three categories:

Metals: They are called as metals because they loose electrons to attain stability. These elements belong from Group 1 to group 13 of the periodic table. They are hard and are mainly solid.
Non-metals: They are called as non-metals because they gain electrons to attain stability. These elements belong from Group 15 to group 17 of the periodic table. They are soft and are mainly liquid or gases.
Metalloids: They are called as metalloids because they show properties of both metals and non-metals. These elements belong from Group 14 of the periodic table.

Oxygen belongs to Group 16 and period 2 of the periodic table.

Hence, oxygen is a non-metal.

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3.0 L of oxygen gas is heated from 100k to 250k. What is the new volume after the temperature has been increased???

alexandr402 [8]

Answer:

7.5 L

Explanation:

Using Charles' law, which is V1/T1=V2/T2, we can plug in these numbers to find the answer. The law states that volume is directly proportional to temperature. 3.0L/100K = x L / 250 K. Solve for x to get 7.5 L. Hope this helps.

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3 years ago

What is the normality of a solution containing 14.8 g of Ca(OH)₂ in 250.0 mL?

aliina [53]

Answer:

Explanation:Are You From Milo?

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3 years ago

A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) C

Mamont248 [21]

Answer:

$T_{eq}=23.85^oC$

Explanation:

Hello,

In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

$\Delta H_{Cu}=-\Delta H_{H_2O}$

Therefore the equilibrium temperature shows up as:

$m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\$

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

$T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC$

Best regards.

6 0

3 years ago

Read 2 more answers

Consider the insoluble compound nickel(II) hydroxide , Ni(OH)2 . The nickel ion also forms a complex with cyanide ions . Write a

natali 33 [55]

Answer: Equilibrium constant for this reaction is $2.8 \times 10^{15}$ .

Explanation:

Chemical reaction equation for the formation of nickel cyanide complex is as follows.

$Ni(OH)_{2}(s) + 4CN^{-}(aq) \rightleftharpoons [Ni(CN)_{4}^{2-}](aq) + 2OH^{-}(aq)$

We know that,

K = $K_{f} \times K_{sp}$

We are given that, $K_{f} = 1.0 \times 10^{31}$

and, $K_{sp} = 2.8 \times 10^{-16}$

Hence, we will calculate the value of K as follows.

K = $K_{f} \times K_{sp}$

K = $(1.0 \times 10^{31}) \times (2.8 \times 10^{-16})$

= $2.8 \times 10^{15}$

Thus, we can conclude that equilibrium constant for this reaction is $2.8 \times 10^{15}$ .

4 0

3 years ago

How many mg of salt are in a 2.78 kg bag of salt?

igomit [66]

Answer:

2,780,000mg

Explanation:

Using the Metric Staircase photo provided, you can calculate how many mg there are in 2.78 kg by simply moving the decimal point six places to the right, since the "Kilo" step takes six places to move to the "Milli" step.

2.78 kg

---------------

2 7 8 0 0 0 0 .

we can see that the decimal point is moved to the right six places.

Answer:

2,780,000mg

7 0

3 years ago