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3 years ago

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1. NaOH mass of a solution of 200g in which its percentage is 25%. What mass of sulfuric acid solution is needed to completely n

eutralize the alkali. What mass of salt will be formed?

1 answer:

Ede4ka [16]3 years ago

4 0

Answer:

$m_{H2SO4$ = 61.25 g

$m_{Na2SO4}$ = 88.75 g

Explanation:

$m_{NaOH}$ = $\frac{200 . 25 }{100}$ = 50 g

⇒ $n_{NaOH}$ = $\frac{50}{40}$ = 1.25 (moles)

2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O

2 : 1 : 1 : 2

1.25 (moles)

⇒ $n_{H2SO4}$ = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ $m_{H2SO4}$ = 0.625 × 98 = 61.25 g

$n_{Na2SO4}$ = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ $m_{Na2SO4}$ = 0.625 × 142 = 88.75 g

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The combustion of glucose (c6h12o6) with oxygen gas produces carbon dioxide and water. this process releases 2803 kj per mole of

V125BC [204]

Answer:- 335 kcal of heat energy is produced.

Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:

$C_6H_1_2O_6+6O_2\rightarrow 6CO_2+6H_2O$

From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.

We could solve this using dimensional analysis as:

$3.00mol O_2(\frac{1mol glucose}{6mol O_2})(\frac{2803 kJ}{1mol glucose})$

= 1401.5 kJ

Now, let's convert kJ to kcal.

We know that, 1kcal = 4.184kJ

So, $1401.5kJ(\frac{1kcal}{4.184kJ})$

= 335 kcal

Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.

8 0

3 years ago

An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr

Zina [86]

<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

$C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa$

Rate of flow of ideal gas , n = 4 kmol/hr = $\frac{4\times 1000mol}{3600s}=1.11mol/s$ (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)

We know that:

$\Delta U=0$ (For isothermal process)

So, by applying first law of thermodynamics:

$\Delta U=\Delta q-\Delta W$

$\Delta q=\Delta W$ .......(1)

Now, calculating the work done for isothermal process, we use the equation:

$\Delta W=nRT\ln (\frac{P_1}{P_2})$

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = final pressure = 50 kPa

Putting values in above equation, we get:

$\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW$

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

$\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW$

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

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What is a surface that reflects a color of 500 nanometer wavelength?

Vinvika [58]

Answer:

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ludmilkaskok [199]

Answer: They use other organisms for energy

Explanation:

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What is the molarity of a solution that contains 2 moles of solute in 4 liters of solution?

Nitella [24]

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n = mols
C = Concentration or Molarity
V = Volume in Liters.

n = 2
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3 years ago