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2 years ago

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

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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

..................(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

substituating the values in eq. (1)

Δx =4.49*10^{-8}10

−8

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

* (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 $HCl$ solution =0.100 moles of HCl

then 0.05L of HCl solution= 0.05 $\times$ 0.1 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 $\times$ 0.0100 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