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Tems11 [23]
2 years ago
5

Write the empirical formula for at least four ionic compounds that could be formed from the following ions:

Chemistry
1 answer:
SpyIntel [72]2 years ago
4 0

The empirical formula of compounds formed from the given ions are as follows:

  • Pb⁴⁺ = PbO₂
  • NH₄⁺ = NH₄Cl
  • CrO₄²⁻ = Na₂CrO₄
  • SO₄²⁻ = K₂SO₄

<h3>What is the empirical formula of a compound?</h3>

The empirical formula of a compound is the simplest formula of the compound showing the simplest ratios in which elements in the compound combine.

The empirical formula of compounds formed from the given ions are as follows:

  • Pb⁴⁺ = PbO₂
  • NH₄⁺ = NH₄Cl
  • CrO₄²⁻ = Na₂CrO₄
  • SO₄²⁻ = K₂SO₄

In conclusion, the empirical formula is the simplest formula of a compound.

Learn more about empirical formula at: brainly.com/question/1581269

#SPJ1

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Will mark as brainlest!
USPshnik [31]

A frequency of 60 MHz is close to the lower end of the old VHF-TV band.


c = f λ ...... where c is the speed of light, f is the frequency and λ is the wavelength

λ = c / f = 3.00x10^8 m/s / 6.0x10^7 1/s

λ = 5.0 m

3 0
3 years ago
Read 2 more answers
a 125 g chunk of aluminum at 182 degrees Celsius was added to a bucket filled with 365 g of water at 22.0 degrees Celsius. Ignor
Diano4ka-milaya [45]
<h3>Answer:</h3>

32.98°C

<h3>Explanation:</h3>

We are given the following;

Mass of Aluminium as 125 g

Initial temperature of Aluminium as 182°C

Mass of water as 265 g

Initial temperature of water as 22°C

We are required to calculate the final temperature of the two compounds;

First, we need to know the specific heat capacity of each;

Specific heat capacity of Aluminium is 0.9 J/g°C

Specific heat capacity of water is 4.184 J/g°C

<h3>Step 1: Calculate the Quantity of heat gained by water.</h3>

Assuming the final temperature is X°C

we know, Q = mcΔT

Change in temperature, ΔT = (X-22)°C

therefore;

Q = 365 g × 4.184 J/g°C × (X-22)°C

    = (1527.16X-33,597.52) Joules

<h3>Step 2: Calculate the quantity of heat released by Aluminium </h3>

Using the final temperature, X°C

Change in temperature, ΔT = -(X°- 182°)C (negative because heat was lost)

Therefore;

Q = 125 g × 0.90 J/g°C × (182°-X°)C

  = (20,475- 112.5X) Joules

<h3>Step 3: Calculating the final temperature</h3>

We need to know that the heat released by aluminium is equal to heat absorbed by water.

Therefore;

(20,475- 112.5X) Joules = (1527.16X-33,597.52) Joules

Combining the like terms;

1639.66X = 54072.52

             X = 32.978°C

                = 32.98°C

Therefore, the final temperature of the two compounds will be 32.98°C

7 0
3 years ago
An atom of argon has a radius rar = 88 pm and an average speed in the gas phase at 25°C of 172 m/s.
Rudik [331]

Answer:

1.2* 10³ rNe.

Explanation:

Given speed of neon=350 m/s

Un-certainity in speed= (0.01/100) *350 =0.035 m/s

As per heisenberg uncertainity principle

Δx*mΔv ≥\frac{h}{4\pi }

4π

h

..................(1)

mass of neon atom =\frac{20*10^{-3} }{6.22*10^{-23} } =3.35*10^{-26} kg

6.22∗10

−23

20∗10

−3

=3.35∗10

−26

kg

substituating the values in eq. (1)

Δx =4.49*10^{-8}10

−8

m

In terms of rNe i.e 38 pm= 38*10^{-12}10

−12

Δx=\frac{4.49*10^{-8} }{38*10^{-12} }

38∗10

−12

4.49∗10

−8

=0.118*10^{4}10

4

* (rNe)

=1.18*10³ rN

= 1.2* 10³ rNe.

Explanation:

This is the answer

7 0
3 years ago
What volume of water is required to prepare 0.1 M H3PO4 from 100 ml of 0.5 M solution?
ExtremeBDS [4]

Answer: A volume of 500 mL water is required to prepare 0.1 M H_{3}PO_{4} from 100 ml of 0.5 M solution.

Explanation:

Given: M_{1} = 0.1 M,    V_{1} = ?

M_{2} = 0.5 M,       V_{2} = 100 mL

Formula used to calculate the volume of water is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\0.1 M \times V_{1} = 0.5 M \times 100 mL\\V_{1} = 500 mL

Thus, we can conclude that a volume of 500 mL water is required to prepare 0.1 M H_{3}PO_{4} from 100 ml of 0.5 M solution.

7 0
3 years ago
A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are
vaieri [72.5K]

Answer: Resulting solution will not be neutral because the moles of OH^-ions is greater. The remaining concentration of [OH^-]ions =0.0058 M.

Explanation:

Given,

[HCl]=0.100 M

[HNO_3] = 0.200 M

[Ca(OH)_2] =0.0100 M

[RbOH] =0.100 M

Few steps are involved:

Step 1: Calculating the total moles of H^+ ion from both the acids

moles of H^+ in HCl

HCl\rightarrow {H^+}+Cl^-

if 1 L of HClsolution =0.100 moles of HCl

then 0.05L of HCl solution= 0.05 \times0.1 moles= 0.005 moles    (1L=1000mL)

moles of H^+ in HCl = 0.005 moles

Similarliy

moles of H^+ in HNO_3

HNO_3\rightarrow H^++NO_3^-}

If 1L of HNO_3 solution= 0.200 moles

Then 0.1L of HNO_3 solution= 0.1 \times 0.200 moles= 0.02 moles

moles of H^+ in HNO_3 =0.02 moles

so, Total moles of H^+ ions  = 0.005+0.02= 0.025 moles     .....(1)

Step 2: Calculating the total moles of [OH^-] ion from both the bases

Moles of OH^-\text{ in }Ca(OH)_2

Ca(OH)_2\rightarrow Ca^2{+}+2OH^-

1 L of Ca(OH)_2= 0.0100 moles

Then in 0.5 L Ca(OH)_2 solution = 0.5 \times0.0100 moles = 0.005 moles

Ca(OH)_2 produces two moles of OH^- ions

moles of OH^- = 0.005 \times 2= 0.01 moles

Moles of OH^- in RbOH

RbOH\rightarrow Rb^++OH^-

1 L of RbOH= 0.100 moles

then 0.2 [RbOH] solution= 0.2 \times 0.100 moles = 0.02 moles

Moles of OH^- = 0.02 moles

so,Total moles of OH^- ions = 0.01 + 0.02=0.030 moles      ....(2)

Step 3: Comparing the moles of both H^+\text{ and }OH^- ions

One mole of H^+ ions will combine with one mole of OH^- ions, so

Total moles of H^+ ions  = 0.005+0.02= 0.025 moles....(1)

Total moles of OH^- ions = 0.01 + 0.02=0.030 moles.....(2)

For a solution to be neutral, we have

Total moles of H^+ ions = total moles of OH^- ions

0.025 moles H^+ will neutralize the 0.025 moles of OH^-

Moles of OH^- ions is in excess        (from 1 and 2)

The remaining moles of OH^- will be = 0.030 - 0.025 = 0.005 moles

So,The resulting solution will not be neutral.

Remaining Concentration of OH^- ions = \frac{\text{Moles remaining}}{\text{Total volume}}

[OH^-]=\frac{0.005}{0.85}=0.0058M

6 0
3 years ago
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