Interaction in which one organism kills and eats another

How many electrons are transferred when magnesium bonds ionically with oxygen

andrew11 [14]

Answer:

2 electrons from magnesium (Mg) is transferred to oxygen (O2) to form Mg2+ and O2-

which combines by strong electrostatic force to form MgO

the equation is

2Mg + O2 -> 2MgO

6 0

3 years ago

Give the formula of the conjugate base of each of the following. (Type your answer using the format CO2 for CO2, (NH4)2CO3 for (

MissTica

Answer:

a) [NH2]-,

b) [O]2-

c) [Cl]-

Explanation:

The conjugate base is the base member (X⁻) of a weak acid (HX). In other words, the conjugate base is the remaining substance due to the proton loss in the acid HX.

a) For NH₃, (HX; X: NH₂⁻), conjugate base is NH₂⁻. In the format, [NH2]-.

b) For OH⁻, (HX; X: O²⁻), conjugate base is O²⁻. In the format, [O]2-.

c) For HCl, (HX; X: Cl⁻), conjugate base is Cl⁻. In the format, [Cl]-.

7 0

3 years ago

A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.

Alexxandr [17]

Answer: The empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

Explanation:

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles$

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 0.078 moles.

For Carbon = $\frac{0.313}{0.078}=4.01\approx 4$

For Hydrogen = $\frac{0.316}{0.078}=4.05\approx 4$

For Oxygen = $\frac{0.078}{0.078}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is $C_4H_4O$

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 = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

$n=\frac{130g/mol}{68g/mol}=1.9\approx 2$

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

$C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2$

Hence, the empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

4 0

3 years ago

At room temperature, a melting solid produces a whitish liquid. A laser beam shined through the liquid is unaffected and produce

lisabon 2012 [21]

The liquid is transparent

3 0

3 years ago

What is the (OH-) in a solution with a pOH of 6.48

inn [45]

Through manipulation of equations, we are able to obtain the equation:

$-pOH= log [ OH^{-}]$

Then we can transform the equation into:

$[ OH^{-}]= 10^{-pOH}$

Then we are able to plug in the pOH and directly get [OH-]:

$[ OH^{-}] = 10^{-6.48}$

$[ OH^{-}]=3.31* 10^{-7} M$

3 0

3 years ago