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xxTIMURxx [149]
3 years ago
12

Which statements are correct about a group? The elements in each group have similar chemical properties. The elements in each gr

oup have identical physical properties. The atomic number of each element increase by 7 down through the group. Each member of the group has the same number of electrons in the outer shell.
Chemistry
2 answers:
yarga [219]3 years ago
8 0
In Chemistry, when we say Group, this term refers to the column in the periodic table wherein the elements in it share the same valence electron structure, as well as their physical and chemical properties. Therefore, based on the statements above, I can say that the answer would be the first option. 
REY [17]3 years ago
8 0

Answer: A and D hope this helps

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Which will result in positive buoyancy and cause the object to float?
Darina [25.2K]
I would say D. Let me know if i am wrong.
3 0
3 years ago
Read 2 more answers
A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2
a_sh-v [17]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

<u>Explanation:</u>

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=13.00g

Mass of H_2O=2.662g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 13.00 g of carbon dioxide, \frac{12}{44}\times 13.00=3.54g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in 2.662 g of water, \frac{2}{18}\times 2.662=0.296g of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles

  • <u>Step 2:</u> 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 0.295 moles.

For Carbon = \frac{0.295}{0.295}=1

For Hydrogen = \frac{0.296}{0.295}=1

For Oxygen = \frac{0.603}{0.295}=2.044\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is CHO_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

n=\frac{90.04g/mol}{45g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4

Hence, the empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

3 0
3 years ago
How many particles are present in0.24moles of carbon?
Ray Of Light [21]

Answer:

1.45 x 10²³ particles

Explanation:

Given parameters:

Number of moles of carbon  = 0.24moles

Unknown:

Number of particles = ?

Solution:

A mole of a substance contains the Avogadro's number of particles.

 The Avogadro's number of particles is 6.02 x 10²³

So;

  0.24 moles of carbon will contain 0.24 x 6.02 x 10²³  = 1.45 x 10²³ particles

7 0
2 years ago
How many Grams of NO is produced if 12g of O2 is combined with excess ammonia?
stich3 [128]

Answer:

9g

Explanation:

moles O2 = mass / Mr = 12 / 2(16.0) = 0.375

ratio O2 : NO = 5:4

moles NO produced = 0.375 * 4/5 = 0.3

mass NO = Mr * mol = (14.0+16.0) * 0.3 = 9g

5 0
2 years ago
Question List (4 items) (Drag and drop into the appropriate area) Find the volume of HCl that will neutralize the base. Find the
expeople1 [14]

The question is incomplete, the complete question is:

The solubility of slaked lime, Ca(OH)_2, in water is 0.185 g/100 ml. You will need to calculate the volume of 2.50\times 10^{-3}M HCl needed to neutralize 14.5 mL of a saturated

<u>Answer:</u> The volume of HCl required is 290mL, the mass of Ca(OH)_2 is 0.0268g, the moles of

<u>Explanation:</u>

Given values:

Solubility of Ca(OH)_2 = 0.185 g/100 mL

Volume of Ca(OH)_2 = 14.5 mL

Using unitary method:

In 100 mL, the mass of Ca(OH)_2 present is 0.185 g

So, in 14.5mL. the mass of Ca(OH)_2 present will be =\frac{0.185}{100}\times 14.5=0.0268g

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

Given mass of Ca(OH)_2 = 0.0268 g

Molar mass of Ca(OH)_2 = 74 g/mol

Plugging values in equation 1:

\text{Moles of }Ca(OH)_2=\frac{0.0268g}{74g/mol}=0.000362 mol

Moles of OH^- present = (2\times 0.000362)=0.000724mol

The chemical equation for the neutralization of calcium hydroxide and HCl follows:

Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O

By the stoichiometry of the reaction:

Moles of OH^- = Moles of H^+ = 0.000724 mol

The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}} .....(2)

Moles of HCl = 0.000724 mol

Molarity of HCl = 2.50\times 10^{-3}

Putting values in equation 2, we get:

2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL

Hence, the volume of HCl required is 290mL, the mass of Ca(OH)_2 is 0.0268g, the moles of

5 0
3 years ago
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