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Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass %

11Alexandr11 [23.1K]

Answer:

For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:

A. 1:1

B. 3:2

C. 2:1

D. 5:2

Note: The question is stated more clearly below:

Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.

What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium?

Explanation:

Number of moles in 100 g mass = % mass / molar mass

Molar mass of Vanadium, V = 51 g/mol

Molar mass of oxygen atom, O = 16 g/mol

1. Percentage mass of V and O is 76.10% and 23.90% respectively.

Number of moles of each atom;

V = 76.10/51.0 = 1.5 moles

O = 23.9/16 = 1.5 moles

Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1

2. Percentage mass of V and O is 67.98% and 32.02% respectively

Number of moles of each atom:

V = 67.98/51 = 1.33

O = 32.02/16 = 2

Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2

3. Percentage mass of V and O is 61.42% and 38.58% respectively

Number of moles of each atom:

V = 61.42/51 = 1.2

O = 38.58/16 = 2.4

Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1

4. Percentage mass of V and O is 56.02% and 43.98% respectively

Number of moles of each atom:

V = 56.02/51 = 1.10

O = 43.98/16 = 2.75

Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2

6 0

3 years ago

Which represents the balanced equation for the beta minus emission of phosphorus-32?

Agata [3.3K]

Answer: The correct option is C.

Explanation: Beta minus ( $\beta ^-$ ) decay is the decay in which a neutron is converted into proton and releases beta particle or generally known as electron.

There is an increase in the Atomic number because one protons is being formed.

In the case of beta-minus decay of phosphorous, the P-nucleus gets changed to Sulfur nucleus. The balanced Chemical equation for this process is:

$^{32}_{15}\textrm{P}\rightarrow ^{32}_{16}\textrm{S}+^0_{-1}\beta$

Hence, the correct option is C.

8 0

3 years ago

Read 2 more answers

Which of these statements describes this mathematical equation?

salantis [7]

Answer:

The final pressure of a gas is inversely proportional to the volume change and directly proportional to temperature

Explanation:

Given

$\frac{P_1V_1}{T_1} = k$

Required

Interpret

$\frac{P_1V_1}{T_1} = k$

Multiply both sides by T1

$T_1 * \frac{P_1V_1}{T_1} = k * T_1$

$P_1V_1 = kT_1$

Divide both sides by V1

$\frac{P_1V_1}{V_1} = \frac{kT_1}{V_1}$

$P_1 = \frac{kT_1}{V_1}$

This can be rewritten as:

$P_1 = k\frac{T_1}{V_1}$

In the above expression; k is a constant of proportionality.

So, the equation can be written as variation as follows:

$P_1\ \alpha\ \ \frac{T_1}{V_1}$

To interpret:

P varies directly to T (the numerator) and inversely to V (the denominator).

3 0

3 years ago

sing any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0°C for the following reaction

prisoha [69]

Answer:

2.76 × 10⁻¹¹

Explanation:

I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.

1. Calculate the free energy of formation of CCl₄

C(s)+ 2Cl₂(g)→ CCl₄(g)

ΔG°/ mol·L⁻¹: 0 0 -65.3

ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹

2. Calculate K

$\text{The relationship between $\Delta G^{\circ}$ and K is}\\\Delta G^{\circ} = -RT \ln K$

T = (25.0 + 273.15) K = 298.15 K

$\begin{array}{rcl}-65 300 & = & -8.314 \times 298.15 \ln K \\65300& = & 2479 \ln K\\26.34 & = & \ln K\\K& = & e^{26.34}\\&= & \mathbf{2.76 \times 10}^{\mathbf{11}}\\\end{array}$

3 0

3 years ago

Determine whether an aqueous solution of each of the following salts will be acidic, basic, or neutral. Ba3(PO4)2 NaCN

Andru [333]

Answer: Both the salts are basic salts.

Explanation:

Salts are formed when an acid reacts with a base during a neutralization reaction.

When a strong acid and a weak base reacts, it leads to the formation of acidic salt.

When a strong base and weak acid reacts, it leads to the formation of basic salt.

When a strong acid and strong base or weak acid and weak base reacts, it leads to the formation of neutral salts.

$Ba_3(PO_4)_2$ is a salt which is formed by the combination of $H_3PO_4\text{ and }Ba(OH)_2$ . Phosphoric acid is a weak acid and barium hydroxide is a strong base. So, the given salt is a basic salt.

$NaCN$ is a salt which is formed by the combination of $HCN\text{ and }NaOH$ . Hydrogen cyanide is a weak acid and sodium hydroxide is a strong base. So, the given salt is a basic salt.

Hence, both the salts are basic salts.

6 0

4 years ago