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3 years ago

Calcium bicarbonate (Ca(HCO3)2) Ca:H:C:O = 1:2:2 _ Lithium sulfide (Li:2s) Li:S = 2:

Chemistry

1 answer:

Anastaziya [24]3 years ago

8 0

Answer: The ratio of atoms in calcium bicarbonate ; Ca : H : C : O = 1:2:2:6.

The ratio of atoms in lithium sulfide; Li : S = 2 : 1

Explanation:

In calcium bicarbonate: $Ca(HCO_3)_2$

In a molecular formula of calcium carbonate there are:

Number of Calcium atoms = 1

Number of Hydrogen atom = 1 × 2 = 2

Number of Carbon atoms = 1 × 2 = 2

Number of Oxygen atoms = 3 × 2 = 6

So, Ca : H : C : O = 1 : 2 : 2 : 6

In lithium sulfide : $Li_2S$

In a molecular formula of lithium sulfide there are:

Number of Lithium atoms = 1 × 2 = 2

Number of Sulfur atoms = 1

So, the Li : S = 2 : 1

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Hello this is my 3 attempt, please someone answer this i'm begging you.

denis23 [38]

Answer:

Explanation:

No worries, I got you :)

So for the first question, you need to use PV=nRT to find the n, or in other words, the number of moles. Then, you can find the molar mass since you know the grams and the moles

110 kPa / 101.3 = 1.085 atm (I converted it to atm so I can use the .08206 L atm/ k mol for the rate)

550 ml / 1000 = .550 L (I converted mL to L in order to use the .08206 L atm/ k mol for the Rate)

28.5 c + 273 = 301.5 K (I converted C to K in order to use the .08206 L atm/ k mol for the Rate)

PV=nRT

(1.085) (.550 L) = n (0.08206) (301.5)

Divide the (0.08206) (301.5) to get n alone:

(1.085) (.550 L) / (0.08206) (301.5) = n

When I divided, I got n= .02412 moles, and since we have 1.88g , we divide the 1.88 by .02412 to get the molar mass (grams/mole)

77.94 g/mole is the molar mass

we know that there are 3 H's in the compound, so we do 3(1.008) and subtract 77.94 by what you get.

3 x 1.008 = 3.024 -----> 77.94-3.024 = 74.9

Now we look at the periodic table and try to find an element that has a molar mass of 74.9

Arsenic (As) has a molar mass of 74.922, which is close enough. Plus, Arsenic has a charge of 3, so it fits with the 3 hydrogens.

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2 years ago

50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH sol

boyakko [2]

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid]) (Eq. 01)

Where

pKa = -log(Ka) (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get

pH = 3.35

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Read 2 more answers

Calcium bicarbonate (Ca(HCO3)2) Ca:H:C:O = 1:2:2 ___ Lithium sulfide (Li:2s) Li:S = 2:__

Calcium bicarbonate (Ca(HCO3)2) Ca:H:C:O = 1:2:2 _ Lithium sulfide (Li:2s) Li:S = 2: