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1 year ago

What is the molarity of a solution made by dissolving 68.4g of NaOH in enough water to produce a 875 ml solution?

1 answer:

8 0

We first calculate for the number of moles of NaOH by dividing the given mass by the molar mass of NaOH which is equal to 40 g/mol. Solving,
moles of NaOH = (68.4 g/ 40 g/mol) = 1.71 moles NaOH
Then, we divide the calculate number of moles by the volume in liters.
molarity = (1.71 moles NaOH / 0.875 L solution)
molarity = 1.95 M

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El amoniaco y el fluor reaccionan para formar tetrafluoruro de dinitrogeno y fluoruro de hidrogeno. segun la reaccion: NH3 + F2

White raven [17]

Answer:

son 12.6 gramos de HF

Explanation:

Tienes que saber qual es el reactor limitante en este caso es fluoruro con los 20 gramos puedes producer .631 mol qual son 12.6 gramos

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2 years ago

A student has two compounds in two separate bottles but with no labels on either one. One is an unbranched alkane, octane, C8H18

Aleks [24]

Answer:

a) both substances are insoluble in water

b) both substances are soluble in ligroin

c) both substances suffer combustion, octane produces more CO₂ than hexene.

d) both substances are less dense than waterl, with hexene having the lowest density.

e) only hexene would react with bromine

f) only hexene would react with permanganate

Explanation:

a) both substances are non-polar and water is polar

b) both substances are non-polar and lingroin is non-polar

c) C₈H₁₈ + 17.5O₂ → 8CO₂ + 9H₂O

C₆H₁₂ + 9O₂ → 6CO₂ + 6H₂O

d) water = 997 kg/m³

ocatne = 703 kg/m³

hexene = 673 kg/m³

e) bromine test is used to detect unsaturations

f) permanganate test is used to detect unsaturations

7 0

2 years ago

For the nitrogen fixation reaction, 3h2(g) + n2(g) 2nh3(g), kc = 6.0 × 10–2 at 500°c. if 0.250 m h2 and 0.050 m nh3 are present

yuradex [85]

Write the Kc equation...
Substitute the given values in the equation and u qill ontain the answer... Give it a try

5 0

1 year ago

Read 2 more answers

The following reaction has Kc > 1: Ni(H2O)62+(aq) + 6 NH3(aq) → Ni(NH3)62+(aq) + 6 H2O(l) Based on this information, what can

Blababa [14]

Answer : The correct option is, $NH_3$ is a stronger Lewis base than $H_2O$

Explanation :

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

As we are given that :

Kc > 1 this means that the reaction is favored in forward direction. That means,

$NH_3$ can readily lose electrons and act as a lewis base which has excess of electrons and a stronger base.

$H_2O$ can readily accept electrons and act as a lewis acid.

Hence, the correct option is, $NH_3$ is a stronger Lewis base than $H_2O$

3 0

2 years ago

2.a. 3.26 g of iron powder are added to 80.0 cm3 of 0.200 mol dm-3 copper(II)

Ksenya-84 [330]

Answer:

The limiting reactant is CuSO₄.

Explanation:

The reaction is:

Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s) (1)

To find the limiting reactant we need to find the number of moles of the reactants.

$\eta_{Fe} = \frac{m}{A_{r}}$

Where:

m: is the mass of iron = 3.26 g

$A_{r}$ : is the standard atomic weight of iron = 55.845 g/mol

$\eta_{Fe} = \frac{3.26 g}{55.845 g/mol} = 0.058 moles$

$\eta_{CuSO_{4}} = M*V$

Where:

M: is the concentration of the CuSO₄ = 0.200 mol/dm⁻³

V: is the volume of the solution = 80.0 cm³

First, we need to convert the units of the volume to dm³ knowing that 1 dm = 10 cm and 1 L= 1 dm³.

$V = 80.0 cm^{3}*\frac{1 dm^{3}}{(10 cm)^{3}} = 0.080 dm^{3}$

Now, the number of CuSO₄ moles is:

$\eta_{CuSO_{4}} = M*V = 0.200 mol/dm^{3}*0.080 dm^{3} = 0.016 moles$

So, to determine the limiting reactant we need to use the molar ratio from equation (1), Fe:CuSO₄ = 1:1

$\eta_{Fe} = \frac{1 mol Fe}{1 mol CuSO_{4}}*0.016 moles \: CuSO_{4} = 0.016 moles$

Since we need 0.016 moles of Fe to react with 0.016 moles of CuSO₄ and initially we have 0.058 moles of Fe, then the limiting reactant is CuSO₄.

Therefore, the limiting reactant is CuSO₄.

I hope it helps you!

8 0

1 year ago