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3 years ago

In the chemical equation Fes - 2HCl → FeCl2 - H2S. there are two FeCl2 molecules

Chemistry

1 answer:

boyakko [2]3 years ago

7 0

Answer:Displacement reaction , as HCl displaces S of FeS to give H2S

Explanation:

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Question 14 A student dissolves 1.5g of styrene C8H8 in 225.mL of a solvent with a density of 1.02/gmL . The student notices tha

Yuki888 [10]

We must to know:

Cm = molarity = niu / Vs, when the niu = no. of moles and Vs = Volume of solution

the no. niu = mass / molecular mass of substance

molecular mass of C8H8 = 12x8+8x1 = 104 g/mol

=> niu = 1,5 / 104 = 0,0144 moles C8H8

=> Cm = 0,0144/0,225 = 0,06 mol/L

Cmm = molality = niu (C8H8) / mass of solvent (kg)

=> p = mass / V => mass (solvent) = p x V

=> 225 x 1,02 = 229,5 g solvent = 0,2295 kg solvent

=> Cmm = 0,0144 / 0,229,5 = 0,063

8 0

3 years ago

Read 2 more answers

What mass of HgO is required to produce 0.692 mol of O2? 2HgO(s) -> 2Hg(l) + O2(g)

Vika [28.1K]

The answer for the following problem is mentioned below.

Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule

Explanation:

Given:

no of moles of the oxygen gas = 0.692

Also given:

2 HgO → 2 Hg + $O_{2}$

where,

HgO represents mercuric oxide

Hg represents mercury

$O_{2}$ represents oxygen

To calculate:

Molar mass of HgO:

Molar mass of HgO = 216 grams

molar mass of mercury (Hg) = 200 grams

molar mass of oxygen (O) =16 grams

HgO = 200 +16 = 216 grams

We know;

2×216 grams of HgO → 1 mole of oxygen molecule

? → 0.692 moles of oxygen molecule

= $\frac{2*216*0.692}{1}$

= 298.944 grams of HgO

Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule

7 0

3 years ago

Read 2 more answers

Select the correct answer.

ser-zykov [4K]

Answer:

I think the answer would be option d.

hope it helps.

3 0

3 years ago

When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2

Arlecino [84]

Answer:

1.62

Explanation:

From the given information:

number of moles of benzamide $=\dfrac{70.4 \ g}{121.14 \ g/mol}$

= 0.58 mole

The molality = $\dfrac{mass \ of \ solute (i.e. \ benzamide )}{mass \ of \ solvent }$

$= \dfrac{0.58 }{0.85 }$

= 0.6837

Using the formula:

$\mathbf {dT = l \times k_f \times m}$

where;

dT = freezing point = 27

l = Van't Hoff factor = 1

kf = freezing constant of the solvent

∴

2.7 °C = 1 × kf × 0.6837 m

kf = 2.7 °C/ 0.6837m

kf = 3.949 °C/m

number of moles of NH4Cl = $\dfrac{70.4 \ g}{53.491 \ g /mol}$

= 1.316 mol

The molality = $\dfrac{1.316 \ mol}{0.85 \ kg}$

= 1.5484

Thus;

the above kf value is used in determining the Van't Hoff factor for NH4Cl

i.e.

9.9 = l × 3.949 × 1.5484 m

$l = \dfrac{9.9}{3.949 \times 1.5484 \ m}$

l = 1.62

5 0

2 years ago

Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)

cupoosta [38]

Answer:

$m_{CaCO_3}=0.179gCaCO_3$

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

$PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2$

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

$m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3$

Regards.

3 0

3 years ago