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4 years ago

Cavalent bonds what are they?

Chemistry

1 answer:

motikmotik4 years ago

5 0

Answer:

A covalent bond, also called a molecular bond, is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.

Explanation:

<h2><u>PLEASE MARK BRAINLIEST!</u></h2>

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Calculate the Molarity (M, moles solute / Liter of solution) of a solution that is made by dissolving 3.456 mmoles of Barium per

yan [13]

Since Molarity is moles per Liter, you will first need to convert the mL into L, which makes it 0.4321L. Next divide: moles on top, Liters on the bottom.
3.456 moles / .4321L = 7.998 M

3 0

3 years ago

How many grams of O2 are present in 44.1 L of O2 at STP?

ycow [4]

Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that occupies.
T is its temperature.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

Then, in this case:

P= 1 atm
V= 44.1 L
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 0°C =273 K

Replacing in the expression for the ideal gas law:

1 atm× 44.1 L= n× 0.082 $\frac{atmL}{molK}$ × 273 K

Solving:

$n=\frac{1 atm x44.1 L}{0.082\frac{atmL}{molK}x273K}$

n=1.97 moles

Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:

$1.97 molesx\frac{32 g}{1 mole}$ = 63.04 g ≈ <u><em>63 g</em></u>

Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

Learn more about the ideal gas law:

<u>brainly.com/question/4147359?referrer=searchResults</u>

7 0

3 years ago

How many moles of hydrochloric acid are required to produce .40 moles of hydrogen gas?

mash [69]

Answer:

0.80 Moles of Hydrochloric Acid would be required to produce 0.40 Moles of hydrogen gas due to the equation ratios

Explanation:

How many moles of hydrochloric acid are required to produce .40 moles of hydrogen gas?

Hydrochloric acid = HCl

Hydrogen gas = H2

H = 1+

Cl = 1-

HCL = H2 + Cl2

2HCl = H2 + Cl2

2:1:1

0.80 = 0.40 + 0.40

4 0

3 years ago

The rate of diffusion of gas A is double the rate of gas B. The ratio of their molecular weight is

Dafna1 [17]

Answer:

Ratio is 4:1

Explanation:

from graham's law of gaseous diffusion:

${ \bf{ \frac{R _{A} }{R_{B} } = \sqrt{ \frac{M_{A} }{M_{B}} } }}$

but Ra = 2Rb:

${ \sf{ (\frac{2}{1} ) {}^{2} = \frac{M _{A}}{M _{B} } }} \\ \\ { \sf{\frac{M _{A}}{M _{B} } = \frac{4}{1} }} \\ \\ {M _{A}} : {M _{B} } = 4 :1$

4 0

3 years ago

Calculation of ph from hydrogen ion concentration what is the ph of a solution that has an h1 concentration of (a) 1.75 3 1025 m

faltersainse [42]

<h2>Solutions:</h2>

<u>Case a:</u> Finding pH for [H⁺] = 1.75 × 10⁻⁵ mol/L :

As we know pH is given as,

pH = -log [H⁺]

Putting value,

pH = -log [1.75 x 10⁻⁵]

pH = 4.75

<u>Case b:</u> Finding pH for [H⁺] = 6.50 × 10⁻¹⁰ mol/L :

As we know pH is given as,

pH = -log [H⁺]

Putting value,

pH = -log [6.50 × 10⁻¹⁰]

pH = 9.18

<u>Case c:</u> Finding pH for [H⁺] = 1.0 × 10⁻⁴ mol/L :

As we know pH is given as,

pH = -log [H⁺]

Putting value,

pH = -log [1.0 × 10⁻⁴]

pH = 4

<u>Case d:</u> Finding pH for [H⁺] = 1.50 × 10⁻⁵ mol/L :

As we know pH is given as,

pH = -log [H⁺]

Putting value,

pH = -log [1.50 × 10⁻⁵]

pH = 4.82

7 0

3 years ago