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3 years ago

What volume of 0.140 HCl is needed to neutralize 2.58 of Mg(OH)2 ? ...?

1 answer:

8 0

To make a first step you have to know the balanced form for neutralization formula: $Mg(OH)2(aq)+2HCl(aq)--\ \textgreater \ 2H2O+Mg(aq)+Cl(aq)
$
According to this, you can <span>calculate what you are being asked :</span> $0.0442molMg(OH)2(x)(2molHCl)/(1molMg(OH)2)=0.0885molHCl$
Then we have : $0.140MHCl=0.140(mol/L)HCl$
Hope everything is clear, here is the exact answer you need : $V=(0.0885molHCl)/(0.140(mol/L)HCl)=0.632LHCl
$

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Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C

ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)

The enthalpy of reaction (1) is given by:

$\Delta H = \Delta H_{r} - \Delta H_{p}$ (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond

7O₂: 7 moles of 1 O=O bond

4CO₂: 4 moles of 2 C=O bonds

6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

$\Delta H = \Delta H_{r} - \Delta H_{p}$

$\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})$

$\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})$

$\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol$

$\Delta H = -2860 kJ/mol$

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

I hope it helps you!

7 0

3 years ago

Naturally occurring gallium is a mixture of isotopes

katrin2010 [14]

It is b have a great rest of your day

7 0

3 years ago

Density is ___ to pressure

Ann [662]

Proportional to pressure

5 0

3 years ago

A 6.1-kg solid sphere, made of metal whose density is 2600 kg/m3, is suspended by a cord. When the sphere is immersed in a liqui

deff fn [24]

Answer:

Density of the liquid = 1470.43 kg/m³

Explanation:

Given:

Mass of solid sphere(m) = 6.1 kg

Density of the metal = 2600 kg/m³

Thus volume of the liquid :

$Volume(V)=\frac{Mass(m)}{Density (\rho)}$

Volume of the sphere = 6.1 kg/2600 kg/m³ = 0.002346 m³

The volume of water displaced is equal to the volume of sphere (Archimedes' principle)

Volume displaced = 0.002346 m³

Buoyant force = $\rho\times gV$

Where

$\rho$ is the density of the fluid

g is the acceleration due to gravity

V is the volume displaced

The free body diagram of the sphere is shown in image.

According to image:

$mg=\rho\times gV+T$

Acceleration due to gravity = 9.81 ms⁻²

Tension force = 26 N

Applying in the equation to find the density of the liquid as:

$6.1\times 9.81=\rho\times 9.81\times 0.002346+26$

$33.841=\rho\times 9.81\times 0.002346$

$\rho=\frac{33.841}{9.81\times 0.002346}$

$\rho=1470.43 kgm^3$

<u>Thus, the density of the liquid = 1470.43 kg/m³</u>

6 0

4 years ago

Which of the following radiations consists of a helium nucleus emitted by some radioactive substances? Alpha radiation Beta radi

erica [24]

Answer: Alpha radiation

Explanation: Alpha decay : When a larger radioactive nuclei decays into smaller nuclei by releasing alpha radiation, the mass number and atomic number is reduced by 4 and 2 units respectively.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}Y+_2^4\textrm{He}$

Beta decay : When a larger radioactive nuclei decays into smaller nuclei by releasing beta radiation, the atomic number is increased by 1 unit.

$_Z^A\textrm{X}\rightarrow _{Z+1}^{A}Y+_{-1}^0e$

Gamma decay : When a larger radioactive nuclei decays into smaller nuclei by releasing gamma radiation, the mass number remains same.

$_Z^A\textrm{X}^*\rightarrow _Z^A\textrm{X}+_0^0\gamma$

5 0

3 years ago