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3 years ago

Which statement is true of the following reaction? H2 + O2 -> 2H+ O2-

Chemistry

1 answer:

Ede4ka [16]3 years ago

6 0

B.it is not balanced for charge or for number of atoms.

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What is the strongest type of intermolecular force between solute and solvent in each solution?

vfiekz [6]

Dipole-Induced-Dipole is the strongest type of intermolecular force between solute and solvent.

<h3>What is Intermolecular Force ?</h3>

Intermolecular force is also called secondary force is the force of attraction between molecules. It acts between ions and atoms.

<h3>What is Dipole-Induced-Dipole attraction ?</h3>

A dipole-induced-dipole attraction is a weak attraction it occurs when the partial charge form with in the molecule due to uneven distribution of charge in a molecule.

CCl₄ is non polar in nature and CH₃OH is polar in nature so dipole-induced-dipole attraction is present.

Thus from the above conclusion we can say that Dipole-Induced-Dipole is the strongest type of intermolecular force between solute and solvent.

Learn more about the Dipole-Induced-Dipole here: brainly.com/question/22973877

#SPJ4

3 0

1 year ago

PLS HELP

AnnyKZ [126]

Answer:

linear ,trigonal planar,tethradral geometry

4 0

2 years ago

How many sulfate ions are there in 321 grams of iron lll sulfate

nikitadnepr [17]

Here, we use the mole as we would use any other collective number: a dozen eggs; a Bakers' dozen; a Botany Bay dozen.
Of course, the mole specifies a much larger quantity, and if I have a mole of stuff then I have
6.022
×
10
23
individual items of that stuff. We can also specify an equivalent mass, because we also know the mass of a mole of iron, and a mole of oxygen etc........The mole is thus the link between the macro world of grams and kilograms and litres, that which we can measure out in the lab, to the micro world of atoms, and molecules, that which we can perceive only indirectly.
Here we have the formula unit
F
e
2
(
S
O
4
)
3
. If there is a mole of formula units, there are necessarily 2 moles of iron atoms, 3 sulfate ions,.......etc.

7 0

2 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

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.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

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

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

Part 1: What is the final volume in milliliters when 0.730 L of a 44.8 % (m/v) solution is diluted to 23.3 % (m/v)?

Andre45 [30]

part 1 : the final volume : 1.404 L

part 2 : the initial concentration : 4.06 M

<h3>Further explanation </h3>

Dilution is the process of adding a solvent to get a more dilute solution.

The moles(n) before and after dilution are the same.

Can be formulated :

M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution

V₁ = volume of the solution before dilution

M₂ = Molarity of the solution after dilution

V₂ = Molarity volume of the solution after dilution

part 1 :

M₁=44.8%

V₁=0.73 L

M₂=23.3%

$\tt V_2=\dfrac{M_1.V_1}{M_2}\\\\V_2=\dfrac{44.8\times 0.73}{23.3}\\\\V_2=1.404~L$

part 2 :

V₁=739 ml=0.739 L

V₂=1.5 L

M₂=2

$\tt M_1=\dfrac{M_2.V_2}{V_1}\\\\M_1=\dfrac{2\times 1.5}{0.739}\\\\M_1=4.06$

6 0

3 years ago