What does ǵene of interest¨ mean ?

When a piece of sodium carbonate reacts with dilute HCl, a gas X is produced. When gas X is passed through lime water then the s

siniylev [52]

Gas x would be carbon dioxide.

note/ acid + carbonate —> salt + water + carbon dioxide

the white precipitate would be calcium carbonate. CaCo₃

note/ this is a common eqn u need to remember.

X - CO₂ (carbón dioxide)
Y - CaCo₃ (calcium carbonate)

sodium carbonate is a basic salt

7 0

3 years ago

Give the formula of each coordination compound. Include square brackets around the coordination complex. Do not include the oxid

rodikova [14]

Answer:

sodium hexachloroplatinate(IV)- Na2[PtCl6]

dibromobis(ethylenediamine)cobalt(III) bromide- [Co(en)2Br2]Br

pentaamminechlorochromium(III) chloride-[Cr(NH3)5Cl]Cl2

Explanation:

The formulas of the various coordination compounds can be written from their names taking cognisance of the metal oxidation state as shown above. The oxidation state of the metal will determine the number of counter ions present in the coordination compound.

The number ligands are shown by subscripts attached to the ligand symbols. Remember that bidentate ligands such as ethylenediamine bonds to the central metal ion via two donors.

4 0

3 years ago

Q #13 How many moles of MgCl2 are there in 350. g of compound?

Vaselesa [24]

<u>Answer:</u>

3.67 moles

<u>Step-by-step explanation:</u>

We need to find out the number of $MgCl_2$ moles present in 350 grams of a compound.

Molar mass of $Mg$ = 24.305

Molar mass of $Cl_2$ = 35.453

So, one mole of $MgCl_2$ = 24.305 + (35.453 * 2) = 95.211g

1 Mole in 1 molecule of $MgCl_2$ = $\frac{1}{95.211} = 0.0105$

Therefore, number of moles in 350 grams of compound = 0.0105 * 350

= 3.67 moles

8 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

EleoNora [17]

Answer:

$\large \boxed{\text{2.20 g Pb}}$

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 239.27 32.00 207.2

2PbS + 3O₂ ⟶ 2Pb + 2SO₃

m/g: 2.54 1.88

2. Calculate the moles of each reactant

$\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}$

3. Calculate the moles of Pb from each reactant

$\textbf{From PbS:}\\\text{Moles of Pb} = \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}$

4. Calculate the mass of Pb

$\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}$

4 0

3 years ago

Find the pH. What are the pH values for the following solutions? (a) 0.1 M HCl (b) 0.1 M NaOH (c) 0.05 M HCl (d) 0.05 M NaOH

slega [8]

Answer:

(a) $pH=1$

(b) $pH=1.3$

(c) $pH=13$

(d) $pH=12.7$

Explanation:

Hello,

In this case, we define the pH in terms of the concentration of hydronium ions as:

$pH=-log([H^+])$

Which is directly computed for the strong hydrochloric acid (consider a complete dissociation which means the concentration of hydronium equals the concentration of acid) in (a) and (c) as shown below:

(a)

$[H^+]=[HCl]=0.1M$

$pH=-log(0.1)=1$

(b)

$[H^+]=[HCl]=0.05M$

$pH=-log(0.05)=1.3$

Nevertheless, for the strong sodium hydroxide, we don't directly compute the pH but the pOH since the concentration of base equals the concentration hydroxyl in the solution:

$[OH^-]=[NaOH]$

$pOH=-log([OH^-])$

$pH=14-pOH$

Thus, we have:

(b)

$pOH=-log(0.1)=1\\pH=14-1=13$

(d)

$pOH=-log(0.05)=1.3\\pH=14-1.3=12.7$

Best regards.

5 0

3 years ago