Under what conditions can calcium bromide conduct electricity

Can you pls tell me the net ironic equation of H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq)+SO²⁻₄(aq) + H₂O(l)

almond37 [142]

Answer:

H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)

Explanation:

H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq) + SO²⁻₄(aq) + H₂O(l)

A careful observation of the equation above, shows that the equation is already balanced.

To obtain the net ionic equation, we simply cancel Mg²⁺ from both side of the equation as shown below:

H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)

5 0

3 years ago

In IR spectroscopy, we normally talk about "frequencies" when in reality we are referring to wavenumbers. What is the mathematic

Svetach [21]

Answer:

Here's what I get.

Explanation:

(b) Wavenumber and wavelength

The wavenumber is the distance over which a cycle repeats, that is, it is the number of waves in a unit distance.

$\bar \nu = \dfrac{1}{\lambda}$

Thus, if λ = 3 µm,

$\bar \nu = \dfrac{1}{3 \times 10^{-6} \text{ m}}= 3.3 \times 10^{5}\text{ m}^{-1} = \textbf{3300 cm}^{-1}$

(a) Wavenumber and frequency

Since

λ = c/f and 1/λ = f/c

the relation between wavenumber and frequency is

$\bar \nu = \mathbf{\dfrac{f}{c}}$

Thus, if f = 90 THz

$\bar \nu = \dfrac{90 \times 10^{12} \text{ s}^{-1}}{3 \times 10^{8} \text{ m$\cdot$ s}^{-1}}= 3 \times 10^{5} \text{ m}^{-1} = \textbf{3000 cm}^{-1}$

(c) Units

(i) Frequency

The units are s⁻¹ or Hz.

(ii) Wavelength

The SI base unit is metres, but infrared wavelengths are usually measured in micrometres (roughly 2.5 µm to 20 µm).

(iii) Wavenumber

The SI base unit is m⁻¹, but infrared wavenumbers are usually measured in cm⁻¹ (roughly 4000 cm⁻¹ to 500 cm⁻¹).

8 0

3 years ago

If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.

vesna_86 [32]

Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

5 0

3 years ago

Read 2 more answers

You are trying to dilute antifreeze to put in your car. You use 1.00 L of the antifreeze with a concentration of 9.24 M. You add

Firlakuza [10]

Use mv=mv, plug in the know 1x9.24=3m. Solve for m, which m is 3.08M

3 0

3 years ago

You have a 15.0 gram sample of gold at 20.0°C. How much heat does it take to raise the temperature to 100.0°C?

Nadusha1986 [10]

Answer:

=154.8 J

Explanation:

The rise in temperature is contributed by the change in temperature.

Change in enthalpy = MC∅, where M is the mass of the substance, C is the specific heat capacity and ∅ is the change in temperature.

Change in temperature = 100.0°C-20.0°C=80°C

ΔH=MC∅

The specific heat capacity of gold= 0.129 J/g°C

ΔH= 15.0g×0.129J/g°C×80°C

=154.8 J

7 0

3 years ago