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2 years ago

Use the drop-down menus to classify the elements as alkali metals, alkaline earth metals, or transition metals.

Chemistry

2 answers:

Rama09 [41]2 years ago

7 0

Nickel (Ni): transition metal

Potassium (K): alkali metal

Cadmium (Cd): transition metal

Calcium (Ca): alkaline earth metal

Barium (Ba): alkaline earth metal

aleksley [76]2 years ago

3 0

The following are the classification of each element above:

Nickel - Transitional Metal
Potassium- Alkali Metal
Cadmium- Transitional Metal
Calcium- Alkaline Earth Metal
Barium- Alkaline Earth Metal

I attached a color-coded, labelled period table that will help to see the derivative of my responses.

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When sodium reacts with chlorine, sodium chloride is produced. Andrew represented this reaction with this equation: Na + Cl2 → 2

My name is Ann [436]

When sodium metal reacts with chlorine gas, the product would be sodium chloride or the table salt. The balanced chemical reaction would be:

2Na + Cl2 = 2NaCl

IN balancing reactions, it is important to remember that the number of atoms at each side should be equal. Hope this answers the question.

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2 years ago

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In a hexagonal-close-pack (hcp) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes is 1:1:2. What i

Alex777 [14]

Explanation:

In a hexagonal-close-pack (HCP) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes = 1 : 1 : 2

let the :

Number of lattice point = 1x.

Number of octahedral points = 1x

Number of tetrahedral points = 2x

If anions occupy the HCP lattice points and cations occupy half of the tetrahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the tetrahedral points, B = $\frac{2x}{2}=x$

The formula of the compound will be = $A_{1x}B_{1x}=AB$

If anions occupy the HCP lattice points and cations occupy all of the tetrahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the tetrahedral points, B = 2x

The formula of the compound will be = $A_{1x}B_{2x}=AB_2$

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2 years ago

Recall that your hypothesis is that these values are the fraction of atoms that are still radioactive after n half-life cycles.

jolli1 [7]

Answer : A= 0.5, B = 0.25 , C = 0.125, D = 0.015625 and E = 0.00390625

Explanation :

Half life of a substance is defined as the amount of time taken by the substance to reduce to half of its original amount.

Here n represents the number of half lives.

The amount of substance that remains after n half lives can be calculated using the given formula, $0.5^{n}$

So when we have n =1,

Fraction of substance that remains = 0.5¹ = 0.5.

That means after first half life over, the amount of substance that remains is 0.5 times that of original.

Therefore we have A = 0.5

When n = 2, we have 0.5² = 0.25

So when 2 half lives are over, the amount of substance that remains is 0.25 times that of original

Therefore B = 0.25

When n = 3, we have 0.5³ = 0.125

So when 3 half lives are over, the amount of substance that remains is 0.125 times that of original.

Therefore we have C = 0.125

When n = 6 , we have 0.5⁶ = 0.015625

So D = 0.015625

When n = 8, we have 0.5⁸ = 0.00390625

Therefore E = 0.00390625

The values for A, B, C, D and E are 0.5, 0.25, 0.125, 0.015625 and 0.00390625 respectively.

8 0

2 years ago

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Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bis

almond37 [142]

Answer: a) $BaCl_2$

b) $BiO_3H_3$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1 : convert given masses into moles.

Moles of Ba = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{66.06g}{137g/mole}=0.48moles$

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ba = $\frac{0.48}{0.48}=1$

For O = $\frac{0.96}{0.48}=2$

The ratio of Ba: Cl= 1:2

Hence the empirical formula is $BaCl_2$

b) Mass of Bi= 80.38 g

Mass of O= 18.46 g

Mass of H = 1.16 g

Step 1 : convert given masses into moles.

Moles of Bi = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{80.38g}{209g/mole}=0.38moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{18.46g}{16g/mole}=1.15moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.16g}{1g/mole}=1.16moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Bi= $\frac{0.38}{0.38}=1$

For O = $\frac{1.15}{0.38}=3$

For H= $\frac{1.16}{0.38}=3$

The ratio of Bi: O: H= 1:3: 3

Hence the empirical formula is $BiO_3H_3$

6 0

2 years ago

Calculate the molarity of sodium chloride in a half-normal saline solution (0.45% NaCl). The molar mass of NaCl is

Rus_ich [418]

Answer:

0.077 M

Explanation:

Data Given :

The concentration of half normal (NaCl) saline = 0.45g / 100 g

So,

Volume of Solution = 100 g = 100 mL

Volume of Solution in Liter = 100 mL / 1000

Volume of Solution = 0.1 L

molar mass of NaCl = 58.44 g/mol

Molarity:

Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M

Formula used for Molarity

M = moles of solute / Liter of solution . . . . . . . . . . (1)

Now to find number of moles of Nacl

no. of moles of NaCl = mass of NaCl / molar mass

no. of moles of NaCl = 0.45g / 58.44 g/mol

no. of moles of NaCl = 0.0077 g

Put values in the eq (1)

M = moles of solute / Liter of solution . . . . . . . . . . (1)

M = 0.0077 g / 0.1 L

M = 0.077 M

So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M

3 0

2 years ago