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3 years ago

A chemist is looking for an element that reacts similarly to the element lithium (LI). Which would be the best choice?

Chemistry

2 answers:

MatroZZZ [7]3 years ago

8 0

Letter d, because they are both alkali metals (group one)

Mama L [17]3 years ago

7 0

Answer: D

Explanation: because i had to same question & got it right

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A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO

Akimi4 [234]

Answer: The empirical formula for the given compound is $CH_2$

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=22.0g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

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

So, in 22.0 g of carbon dioxide, $\frac{12}{44}\times 22.0=6g$ of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles$

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 0.5 moles.

For Carbon = $\frac{0.5}{0.5}=1$

For Hydrogen = $\frac{1.0}{0.5}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is $C_{1}H_{2}=CH_2$

4 0

4 years ago

Which of the following is the least reactive nonmetal? A. I B. F C. Br D. Cl

tatyana61 [14]

the answer is A. I (iodine)

no worries

5 0

3 years ago

Why did most of the alpha particles go straight through the gold foil in Rutherford's experiment?

andre [41]

Answer:

The fact that most alpha particles went straight through the foil is because the atom is mostly empty space.

Those that passed straight through did so because they didn't encounter any nuclei.

Explanation:

4 0

2 years ago

How many moles of agcl are contained in 244 ml of 0.135 magcl solution? the density of the solution is 1.22 g/ml?

spin [16.1K]

Since we are only asked for the number of moles, we don't need the information of density. The concentration is expressed in terms of 0.135 M AgCl or 0.135 moles of AgCl per liter solution. The solution is as follows:

Moles AgCl = Molarity * Volume
Moles AgCl = 0.135 mol/L * 244 mL * 1 L/1000 mL
Moles AgCl = 0.03294 mol 

3 0

3 years ago

Read 2 more answers

State which of the following set of quantum number would be possible and which would be permissible for an electron in an atom.

raketka [301]

Answer:

The correct option is: e) n=4, l=3, $m_{l}$ = - 2 , $m_{s}$ = - 1/2

Explanation:

The four quantum numbers that describe the electrons in an atom are: The principal: n, azimuthal: l, magnetic: m, and spin: s.

For a given electron, the values of the quantum numbers should be in the given range- principal quantum number: n ≥ 1;

azimuthal quantum number: 0 ≤ l ≤ n − 1;

magnetic quantum number: −l ≤ $m_{l}$ ≤ +l;

spin quantum number: −s ≤ $m_{s}$ ≤ +s

Now, for n= 4,

The value of l: 0 to n − 1 = 0 to 3 = 0, 1, 2, 3

The value of $m_{l}$ for l= 3 : -3 to +3 = -3, -2, -1, 0, +1, +2, +3

The value of $m_{s}$ : -1/2, +1/2

Therefore, the given set of quantum numbers: n=4, l=3, $m_{l}$ = - 2 , $m_{s}$ = - 1/2; are possible and permissible for an electron in an atom.

5 0

3 years ago