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tia_tia [17]
3 years ago
6

A laboratory technician combined sodium hydroxide with excess iron(II) nitrate. A reaction took place according to this chemical

equation: 2NaOH + Fe(NO3)2 → NaNO3 + Fe(OH)2.
The reaction produced 3.70 grams of iron(II) hydroxide.

Assuming the reaction came to completion, what was the initial mass of sodium hydroxide? Use the periodic table.

A. 1.6 g
B. 2.0 g
C. 3.3 g
D. 4.0 g
Chemistry
1 answer:
Whitepunk [10]3 years ago
8 0

 The initial  mass  of sodium hydroxide  is  3.3 g (answer C)

 <u><em>calculation</em></u>

Step 1 : find the  moles of iron (ii) hydroxide (  Fe(OH)₂

moles =  mass÷  molar mass

from periodic table the  molar mass of Fe(OH)₂  = 56 + [16 +1]2  = 90 g/mol

moles  is therefore = 3.70 g÷ 90 g/mol = 0.041 moles

Step 2:  use the mole ratio to  calculate the moles of  sodium hydroxide (NaOH)

   from given equation  NaOH : Fe(OH)₂    is 2 :1

therefore the moles of NaOH = 0.041 x 2 = 0.082 moles

Step 3: find  mass of NaOH

mass = moles x molar mass

from the periodic table the  molar mass of NaOH = 23 +16 +1  = 40 g/mol

mass  = 0.082  moles x 40 g/mol = 3.3 g ( answer C)


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The vapor pressure of water at 25.0°c is 23.8 torr. determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.
svp [43]
Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
when we have Pv(solu) = 22.8 and Pv(water) = 23.8 so by substitution:
22.8 = n * 23.8
n= 0.958
- we need to get the moles of glucose:
moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958   = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
0.968 moles of glucose = 1.2
moles of glucose = 1.253 mol
∴ the mass of glucose = no.of glucose moles x molar mass 
                                      = 1.253 x 180 = 225.5 g
Q: B
here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
n = 0.88
so, mole fraction of solution = 1 - 0.88 = 0.12
and we can get after that the moles of water = (mass weight / molar mass)
- no.moles of water = 85 g / 18 g/mol = 4.7 moles
- total moles in solution = moles of water / moles fraction of water 
                                        = 4.7 / 0.88 = 5.34 moles 
∴ moles of the solution = total moles in solu - moles of water 
                                       = 5.34 - 4.7 = 0.64 moles solute
∴ the molar mass of the solute = mass weight of solute / no.of moles of solute
                                                    = 53.8 / 0.64 = 84 g/mole

Q: C

moles of urea (NH2)2 CO = mass weight / molar mass
                                           = 4.49 g / 60 g /mol
                                           = 0.07 mol
moles of methanol = mass weight / molar mass 
                                 = 39.9  g / 32  g/mol = 1.25 mol
moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
moles fraction of methanol = 1.25 / ( 1.25+0.07) = 0.95
by substitution in Pv formula we will be able to get the vapour pressure of the solu :
Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg 
∴Pv(solu) = 84.55 mmHg


 
7 0
3 years ago
All of the following are reasons why equations or balanced EXCEPT
Anni [7]

Answer:

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Explanation:

4 0
2 years ago
For the equilibrium
Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

3 0
2 years ago
Which of the following is the simplest ketone used as an organic solvent?
Lyrx [107]

Answer: Option (d) is the correct answer.

Explanation:

A ketone is an organic functional group that contains a carbon and oxygen atom bonded together through a double bond, that is, C=O.

For example, acetone CH_{3}-CO-CH_{3} is a ketone.

Whereas a hydrocodone is a drug which is used to relieve from moderate to sever pain.It is mostly combined with other drugs and resulting in a chemical formula C_{18}H_{21}NO_{3}.C_{4}H_{6}O_{6}.

A camphor is a volatile white color substance with chemical formula C_{10}H_{16}O. It has aromatic smell and its taste is bitter.

A menthone is also an organic compound with chemical formula C_{10}H_{18}O.

Thus, we can conclude that out of the given options the simplest ketone used as an organic solvent is acetone.

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