At 298 K and 1 atm, bromine is a liquid with a high vapor pressure, whereas chlorine is a gas. This provides evidence that, unde

Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Ca2+, V5+, Br-

lorasvet [2795]

Answer:

CaS, CaBr₂, VBr₅, and V₂S₅.

Explanation:

The ionic compound must exhibit neutrality; its total charge should equal zero.

A binary ionic compound is formed from two distinct ions.

Ca²⁺ combines with either Br⁻ or S²⁻ to create binary ionic compounds.

CaS is created when Ca²⁺ pairs with S²⁻ resulting in the neutral binary ionic compound CaS.

CaBr₂ results from the combination of one mole of Ca²⁺ with two moles of Br⁻ to form the neutral binary ionic compound CaBr₂.

V⁵⁺ can also unite with either Br⁻ or S²⁻ to produce binary ionic compounds.

V₂S₅ is formed when two moles of V⁵⁺ bond with five moles of S²⁻ yielding the neutral binary ionic compound V₂S₅.

VBr₅ is produced by combining one mole of V⁵⁺ with five moles of Br⁻ to form the neutral binary ionic compound VBr₅.

Thus, the empirical formulas for four binary ionic compounds that may be produced are: CaS, CaBr₂, VBr₅, and V₂S₅.

5 0

3 months ago

Compounds A and B are colorless gases obtained by combining sulfur with oxygen. Compound A results from combining 6.00 g of sulf

lorasvet [2795]

Answer:

1.5

Explanation:

It is given that:

Compound A and B originate from Sulfur + Oxygen.

Compound A:

6g sulfur + 5.99g Oxygen

Compound B:

8.6g sulfur + 12.88g oxygen

By comparing the ratios:

Compound A:

S: O = 6.00: 5.99

S/0 = 6.0g S / 5.99g O

Compound B:

S: O = 8.60: 12.88

S / O = 8.60g S / 12.88g O

The mass ratio of A and that of B

(6.0g S / 5.99g O) ÷ (8.60g S / 12.88g O)

(6.0 g S / 5.99g O) × (12.88g O / 8.60g S)

(6 × 12.88) / (5.99 × 8.60)

= 77.28 / 51.514

= 1.50017

= 1.5

4 0

2 months ago

Combustion analysis of an unknown compound containing only carbon and hydrogen produced 0.2845 g of co2 and 0.1451 g of h2o. wha

VMariaS [2998]

CxHy + (x+0.25)O₂ → xCO₂ + 0.5yH₂O

m(CO₂)/{xM(CO₂)}=m(H₂O)/{0.5yM(H₂O)}

0.2845/{44.01x}=0.1451/{9.01y}

x/y=0.4=2:5

The empirical formula is C₂H₅.

7 0

2 months ago

66.667 mL of 3.000 M H2SO4 (aq) solution was neutralized by the stoichiometric amount of 4.000 M Al(OH)3 solution in a coffee cu

eduard [2782]

Answer:

$\large \boxed{\Delta_{\textbf{r}}H =\text{-4600 J$\cdot$ mol}^{-1}}$

Explanation:

This scenario is unrealistic since Al(OH)₃ is not soluble in water.

The question consists of two parts:

A. Stoichiometry — where we determine volumes, masses, and moles for the products

B. Calorimetry — where we assess the enthalpy of the reaction.

A. Stoichiometry

1. Determine the volume of Al(OH)₃

(a) The balanced chemical equation:

2Al(OH)₃ + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 6H₂O

M/V: 66.667

c/mol·L⁻¹: 4.000 3.000

(b) Moles of H₂SO₄

$\rm \text{66.667 mL H$_{2}$}SO_{4} \times \dfrac{\text{3.000 mmol H$_{2}$SO}_{4}}{\text{1 mL H$_{2}$SO}_{4}} = \text{200.00 mmol H$_{2}$SO}_{4}$

(c) Moles of Al(OH)₃

The molar ratio stands at 2 mmol Al(OH)₃: 3 mmol H₂SO₄

$\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H$_{2}$SO}_{4} \times \dfrac{\text{2 mmol Al(OH)}_{3}}{\text{3 mmol H$_{2}$SO}_{4}}\\\\= \text{133.33 mmol Al(OH)}_{3}$

(d) Volume of Al(OH)₃

$\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H$_{2}$SO}_{4} \times \dfrac{\text{1 mL Al(OH)}_{3}}{\text{4 mmol H$_{2}$SO}_{4}} = \text{50.000 mL Al(OH)}_{3}$

B. Calorimetry

This reaction has two energy exchanges.

q₁ = heat from the reaction

q₂ = heat used to heat the calorimeter

q₁ + q₂ = 0

nΔH + mCΔT = 0

Data:

Moles of Al₂(SO₄)₃ = 0.066 667 mol

C = 1.10 J°C⁻¹g⁻¹

T_initial = 22.3 °C

T_final = 24.7 °C

Calculations

(a) Mass of solution

Assume solutions are as dense as water (though not realistic).

Mass of sulfuric acid solution = 66.667 g

Mass of aluminium hydroxide solution = 50.000

TOTAL = 116.667 g

(b) ΔT

ΔT = T_final - T_initial = 24.7 °C - 22.3 °C = 2.4°C

(c) ΔH

$\begin{array}{ccccl}n\Delta H & +& mC \Delta T& = &0\\\text{0.066 667 mol }\times \Delta H& + & \text{116.667 g} \times 1.10 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 2.4 \, ^{\circ}\text{C} & = & 0\\0.066667 \Delta H \text{ mol} & + & \text{310 J} & = & 0\\&&0.066667 \Delta H \text{ mol} & = & \text{-310 J} & & \\\end{array}\\$

$\begin{array}{ccccl}& &\Delta H & = & \dfrac{\text{-310 J}}{\text{0.066667 mol}}\\\\& &\Delta H & = & \textbf{-4600 kJ/mol}\\\end{array}\\\large \boxed{\mathbf{\Delta_{\textbf{r}}H} =\textbf{-4600 J$\cdot$ mol}^{\mathbf{-1}}}$

This result appears nonsensical, but it is derived from your given figures.

6 0

2 months ago

How many micrograms of iron were in the 8.0 mL sample of Greg's blood?

castortr0y [3046]

The result is: 3.36 micrograms of iron in<span> Greg's blood sample.
</span>m(Fe) = 42 mcg(micrograms).
V(Fe) = 1 dL = 1 dL · 100 mL/1dL.
V(Fe) = 100 mL.
Using proportions: m(Fe): 8 mL = 42 mcg: 100 mL.
Thus, 100 mL · m(Fe) = 8 mL · 42 mcg.
m(Fe) = 336 mL·mcg ÷ 100 mL.
m(Fe) = 3.36 mcg.

4 0

2 months ago