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Dafna11 [192]
3 years ago
5

How many moles of NaOH are in 27.8 mL of 0.168 M NaOH?

Chemistry
1 answer:
Alex17521 [72]3 years ago
8 0
Your answer would be 0.00285 moles.
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What is the minimum volume of 5.50 M HCl necessary to neutralize completely the hydroxide in 718.0 mL of 0.183 M NaOH
Deffense [45]

The volume of HCl required is 23.89 mL

Calculation of volume:

The reaction:

HCl + NaOH \rightarrow NaCl + H_2O

As HCl and NaOH react in 1 : 1 ratio.

Volume of NaOH= 718 mL

Concentration= 0.183M

Volume of HCl= ?

Concentration= 5.50M

Using the dilution formula:

M_1\times V_1(NaOH)= M_2\times V_2(HCl)

0.183\times 718= 5.50 \times V_2\\V_2=\frac{131.394}{5.50} \\V_2 = 23.89 \,mL

Therefore,

Volume of HCl required will be 23.89 mL.

Learn more about neutralization reaction here,

brainly.com/question/1822651

#SPJ4

6 0
1 year ago
What is the pressure of 1.50 moles of a gas in a 30.0L tank at a temperature of<br> 285K?
julia-pushkina [17]

The pressure of the gas : 1.1685 atm

<h3>Further explanation</h3>

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm  

V = volume, liter  

n = number of moles  

R = gas constant = 0.08206 L.atm / mol K  

T = temperature, Kelvin  

n=moles=1.5

V=volumes = 30 L

T=temperature=285 K

The pressure :

\tt P=\dfrac{nRT}{V}\\\\n=\dfrac{1.5\times 0.082\times 285}{30}\\\\P=1.1685~atm

7 0
2 years ago
Which of the following elements is the most electronegative?
Veseljchak [2.6K]
Chlorine is the most electronegative
5 0
3 years ago
Read 2 more answers
What is The Magnus Effect/Force? Explain.
cluponka [151]

Answer:

dk

Explanation:

6 0
3 years ago
50g nitrous oxide combines with 50g oxygen form dinitrogen tetroxide according to the balanced equation below.
photoshop1234 [79]

Limiting reactant : O₂

Mass of  N₂O₄ produced = 95.83 g

<h3>Further explanation</h3>

Given

50g nitrous oxide

50g oxygen

Reaction

2N20 + 302 - 2N204

Required

Limiting reactant

mass of N204 produced

Solution

mol N₂O :

\tt =\dfrac{50}{44}=1.136

mol O₂ :

\tt =\dfrac{50}{32}=1.5625

2N₂O+3O₂⇒ 2N₂O₄

ICE method

1.136    1.5625

1.0416  1.5625    1.0416

0.0944    0          1.0416

Limiting reactant : Oxygen-O₂

Mass N₂O₄(MW=92 g/mol) :

\tt =mol\times MW=1.0416\times 92=95.83~g

7 0
3 years ago
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