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3 years ago

A transition metal in the fourth period from the following list : Cu, O , Pr, Ag

Chemistry

1 answer:

Kaylis [27]3 years ago

3 0

Answer:

Explanation:

Groups 3 - 12 (or groups IIA - IIB) of the periodic table contain transition elements. Transaction elements start from period four (4) of the periodic table. The phrase alludes to the fact that the d sublevel is filling at a lower main energy level than the s sublevel that came before it.

The transition elements' arrangement is inverted from the fill order, with the 4 s filled prior to the actual 3 d begins. The transition elements are commonly referred to as transition metals since they are all metals. They are less reactive than the metals in Groups 1 and 2 and have normal metallic characteristics.

From the options given Cu is the only transition metal in the fourth period on the periodic table.

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Does seeing a gas produced always mean a chemical reaction is occurring? Explain.

chubhunter [2.5K]

Seeing signs of a chemical reaction does not always mean that a reaction is happening. For example, a gas (water vapor) is given off when water boils. ... You can tell that it is a physical change because water vapor can condense to form liquid water. In a chemical change, a new substance must be produced.

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3 years ago

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A balloon contains a 10% glucose solution. The balloon is permeable to water but not to glucose. A beaker contains a 5% glucose

deff fn [24]

The state of the balloon:

When the balloon is submerged in the beaker, the amount of water in the beaker will get reduced.

What is Osmosis:

Based on the concentration of solutes on both sides of the membrane, water will flow through a permeable membrane in a specific direction.

Hypertonic solution:

It means that there are more solutes present in the surrounding environment than in the cell itself.

Hypotonic solution:

In a hypotonic solution, the concentration of solutes inside the cell is higher than that outside the cell.

When comparing two solutions, the one with the larger solute concentration is hypertonic, and the one with the lower solute concentration is hypotonic. Isotonic solutions have an identical solute concentration.

While the solution in the beaker is hypertonic, Meaning that will draw water molecules out of the cell. As water molecules move from a location of high water potential (dilute solution) to a region of reduced water potential (10% glucose solution), the water from the 5% glucose solution will flow into the 10% one (concentrated solution)
This is the reason why the amount of water decreases when the balloon is submerged in the beaker.

Learn more about the glucose solution and permeability here,

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2 years ago

Element A has a half-life of 10 days. A scientist measures out 200 g of this substance. After 30 days has passed, the scientist

myrzilka [38]

N=N₀*2^(-t/T)

N₀=200 g
T=10 d
t=30 d

N=200*2^(-30/10)=25 g

25 g will remain

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3 years ago

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In the first distillation this week, which component of the original solvent mixture makes the larger contribution to the vapor

Olenka [21]

In the first distillation this week, Hexane from the original solvent makes a larger contribution to the vapor pressure of the mixture.

In between hexane and toluene, the hexane will have more vapor pressure contribution in the solution. The boiling point of hexane is much lower than toluene. Therefore, it will evaporate easily at low temperatures and start exerting pressure on the solution.

Hence between hexane and toluene, because of more vapor pressure of hexane and lower boiling point, it will easily evaporate and exerts pressure.

Therefore, from the original solvent, hexane makes a larger contribution to the vapor pressure of the mixture.

To learn more about vapor pressure and hexane, visit: brainly.com/question/28206662

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1 year ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

3 years ago