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timurjin [86]
3 years ago
6

Aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . Suppose 1.6

g of hydrobromic acid is mixed with 1.15 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Chemistry
2 answers:
Andreas93 [3]3 years ago
8 0

Answer:

0.35 g of H₂O

Explanation:

Firstly we have to make the reaction's equation.

In this case, the reactants are HBr and NaOH and the products are water and NaBr. This is a neutralization reaction.

We calculate the moles of each reactant:

1.6 g / 80.9 g/mol = 0.0198 moles of HBr

1.15 g / 39 g/mol = 0.0295 moles of NaOH

Reaction: NaOH + HBr →  NaBr + H₂O

Ratio is 1:1, so the limiting reactant is the hydrobromic acid.

For 0.0295 moles of hydroxide we need the same amount of acid, but we do not have enough HBr, just 0.0198 moles.

Ratio is 1:1, again. 1 mol of acid can produce 1 mol of water.

Therefore, 0.0198 moles of HBr will produce 0.0198 moles of water.

We convert the moles to mass → 0.0198 mol . 18 g/mol = 0.35 g of H₂O

Licemer1 [7]3 years ago
4 0

Answer:

0.36 grams of H2O will be produced

Explanation:

Step 1: data given

Mass of hydrobromic acid (HBr) = 1.6 grams

Molar mass HBr = 80.91 g/mol

Mass of sodium hydroxide (NaOH) = 1.15 grams

Molar mass of NaOH = 40.0 g/mol

Step 2: The balanced equation

HBr + NaOH → NaBr + H2O

Step 3: Calculate moles HBr

Moles HBr = mass HBr / molar mass HBr

Moles HBr = 1.6 grams / 80.91 g/mol

Moles HBr = 0.0198 moles

Step 4: Calculate moles NaOH

Moles NaOH = 1.15 grams / 40.0 g/mol

Moles NaOH = 0.0288 moles

Step 5: Calculate limiting reactant

For 1 mol HBr we need 1 mol NaOH to produce 1 mol NaBr and 1 mol H2O

HBr is the limiting reactant. It will completely be consumed (0.0198 moles). NaOH is in excess. There will react 0.019 moles. There will remain 0.0288 - 0.0198 = 0.009 moles

Step 6: Calculate moles H2O

For 1 mol HBr we need 1 mol NaOH to produce 1 mol NaBr and 1 mol H2O

For 0.0198 moles HBr we'll have 0.0198 moles of H2O

Step 7: Calculate mass H2O

Mass H2O = 0.0198 moles * 18.02 g/mol

Mass H2O = 0.36 grams

0.36 grams of H2O will be produced

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. Helium is stored at 293 K and 500 kPa in a 1-cm-thick, 2-m-inner-diamater spherical container made of fused silica. The area w
Ivahew [28]

Answer:  

(a) 45.17×10^-14 kg/s  

(b) since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.  

Explanation:  

Helium gas at temperature T=293k  

Helium gas at pressure P= 500kPa  

The inner diameter of spherical tank is D_1 = 2m  

The inner radius of spherical tank is : r_1 = \frac{D_1}{2}  

= \frac{2}{2}  

=1m  

Thickness of the container r = 1cm =0.01m  

Outer radius of the spherical tank is ;  

t = r_2 - r_1  

-r_2 = -t - r-1  

multiplying through with (-) we have ;  

r_2 = t + r_1  

r_2 = 1 + 0.01m  

r_2 = 1.01m  

From table of binary diffusion coefficients of solids, the diffusion coefficients of helium in silica is noted as  

D_A_B =4.0 ×10^-14 \frac{m^2}{s}  

From table molar mass and gas constant, the molecular weight of helium is:

 

M = 4.003kg/kmol  

The solubility of helium in fused silica is determined from Table of Solubility of selected gases and soilids.  

S_He = 0.00045 kmol/m³. bar  

Considering total molar concentration as constant, the molar concentration of helium inside the container is determined as  

C_B_I = S_H_e×P  

= 0.00045kmol/m³. bar × (5)  

C_B_I = 2.25×10^-3 kmol/m³  

From one dimensional mass transfer through spherical layers is expressed as:

N_di_f_f= 4πr_1 r_2 D_A_B \frac{C_B_I - C_B_2}{r_2 - r_1}

substituting all the values in the above relation, we have;

M_di_f_f= 4π(1) (1.01) (4.0×10^-14) \frac{2.25 × 10^-3 -0}{1.01-1}

M_di_f_f=11.42×10^-14kmol/s

(a) The mass flow rate is expressed as

M_di_f_f = MN_diff

M_di_f_f=4.003×11.42×10^-14

M_di_f_f=45.71×10^-14kg/s

(b) The pressure drop in the tank after a week;

For one week the mass flow rate of helium is

N_di_ff =11.42×10^-14kmol/s

N_di_ff= 11.42×10^-14×7×24×3600 kmol/week

N_di_f_f=6.9×10^-8kmol/week

The volume of the spherical tank is V=\frac{4}{3} πr_1^3

V=\frac{4}{3}π×1^3

V = 4.189m³

The initial mass of helium in the sphere is determined from the ideal gas equation:

PV=NRT

where R is the universal gas constant and its value is R = 8.314 KJ/Kmol.k

N= PV/RT

N= 500 × 4.189/ 8.314 × 293

N= 0.86kmol

The number of moles of helium gas remaining in the tank after one week is:

N_di_f_f-final = 0.86 - 6.9 × 10^-8 kmol/week

N_di_f_f-final ≅ 0.86

therefore, since the amount of helium escaped due to diffusion is insignificant, the final pressure drop in the tank remains the same as the initial pressure 500kPa.  

3 0
3 years ago
Which one of the following characteristics indicates the presence of weak intermolecular forces in a liquid?A) none of the above
grin007 [14]

Answer:

C) a low heat of vaporization.

Explanation:

It is C) a low heat of vaporization.

Low heat of vaporization means less amount of energy is required to vaporize a liquid. This further means the liquid has weaker intermolecular force of attraction that is why on provide lesser heat only it vaporizes.

Low vapor pressure, high boiling point and high critical temp. all contribute to strong intermolecular forces in a liquid

6 0
3 years ago
Question
guapka [62]

Answer:

677

Explanation:

took the test

it said that it was wrong but then it also said the total bond energy of the reactants is 677 kJ/mol. the correct one was 854, i guess.

8 0
3 years ago
Read 2 more answers
How many moles are equal to 12.65 grams of Al2(SO4)3?
____ [38]

Answer:

0.03697 mol Al₂(SO₄)₃

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

12.65 g Al₂(SO₄)₃

<u>Step 2: Identify Conversions</u>

Molar Mass of Al - 26.98 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Al₂(SO₄)₃ - 2(26.98) + 3(32.07) + 12(16.00) = 342.17 g/mol

<u>Step 3: Convert</u>

<u />12.65 \ g \ Al_2(SO_4)_3(\frac{1 \ mol \ Al_2(SO_4)_3}{342.17 \ g \ Al_2(SO_4)_3} ) = 0.03697 mol Al₂(SO₄)₃

<u>Step 4: Check</u>

<em>We are given 4 sig figs. Follow sig fig rules and round.</em>

We already have 4 sig figs in the final answer, so no need to round.

5 0
3 years ago
Select all that apply. Which statements concerning this diagram are correct?
belka [17]
Hello, let me break this down for you! :)
A is your energy of your reactants, B is your activation energy, so the total value could be found from B-Y. C is the value of your energy of your products. X is your delta H value. So in this, your products are higher energy than the reactants, so H is + and e2-E1 is positive so E2 >E1
Therefore your answers would just be <span>E2 > E1 and
</span><span>Hrxn<span> is positive.
Hope this helps! If you have any other questions or need further explanation just let me know! :)
</span></span>
5 0
4 years ago
Read 2 more answers
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