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3 years ago

How many grams of nickel are in a 25.0 g sample of nickel (II) fluoride?

Chemistry

1 answer:

snow_tiger [21]3 years ago

5 0

Answer:

15.17 g

Explanation:

To answer this, we need to find the molar mass of nickel in nickel (II) fluoride. The formula for nickel (II) fluoride is NiF2. This gives us the molar mass of 96.69 g. The mass percentage of nickel is 60.70% approximately (as we divide the molar mass of nickel by that of nickel (II) fluoride), and 60% of 25g gives us 15.17 g

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Enter a balanced equation for the complete combustion of liquid C3H7OH. Express your answer as a chemical equation. Identify all

Vlad1618 [11]

2 C₃H₇OH (l) + 9 O₂ (g) → 6 CO₂ (g) + 8 H₂O (g)

Explanation:

To balance the chemical equation the number of atoms of each element entering the reaction have to be equal to the number of atoms of each element leaving the reaction, in order to conserve the mass.

Bellow we have the balanced chemical equation of the complete combustion of C₃H₇OH:

C₃H₇OH (l) + (9/2) O₂ (g) → 3 CO₂ (g) + 4 H₂O (g)

to have integer coefficients we multiply the reaction with 2:

2 C₃H₇OH (l) + 9 O₂ (g) → 6 CO₂ (g) + 8 H₂O (g)

where:

(l) - liquid

(g) - gaseous

Learn more about:

combustion reaction

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balancing chemical equations

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#learnwithBrainly

7 0

3 years ago

How many Joules must be removed to condense 150.0g of steam at 100.0 degrees Celsius to water at 100.0 degrees celcius

Elina [12.6K]

Answer:

339kJ

Explanation:

Given parameters:

Mass of steam = 150g = 0.15kg

Initial temperature of steam = 100°C

Final temperature of water = 100°C

Unknown:

Quantity of heat that must be removed to condense the steam = ?

Solution:

The heat involved here is a latent heat because there is no change temperature. The process is just a phase change.

H = mL

m is the mass

L is the latent heat of vaporization = 2,260 kJ/kg

Insert the parameters and solve;

H = 0.15kg x 2,260 kJ/kg

H = 339kJ

7 0

3 years ago

What volume (L) of 0.250 M HNO3 is required to neutralize a solution prepared by dissolving 17.5 g of NaOH in 350 mL of water ?

IRINA_888 [86]

<h3>Answer:</h3>

1.75 L HNO₃

<h3>Explanation:</h3>

We are given;

Molarity of HNO₃ as 0.250 M

Mass of NaOH as 17.5 g

Volume of NaOH = 350 mL

We are required to calculate the volume of 0.250 M

We are going to first write the balanced reaction:

NaOH(aq) + HNO₃(aq) → NaNO₃(aq) + H₂O(l)

Then, we calculate the number of moles of NaOH

Moles = Mass ÷ Molar mass

Molar mass of NaOH = 39.997 g/mol

= 17.5 g ÷ 39.997 g/mol

= 0.4375 moles

We can now calculate the number of moles of HNO₃ using the mole ratio from the equation;

Mole ratio of NaOH to HNO₃ is 1 : 1

Therefore, if moles of NaOH are 0.4375 moles then;

Moles of HNO₃ will also be 0.4375 moles

We can now calculate the volume of HNO₃

Morality = Number of moles ÷ Volume

Thus;

Volume = Number of moles ÷ Molarity

= 0.4375 moles ÷ 0.250 M

= 1.75 L

Therefore, the volume of HNO₃ is 1.75 L

5 0

3 years ago

Calculate the concentrations of all species present in a 0.26 M solution of ethylammonium chloride (C2H5NH3Cl).

Alina [70]

Answer:

0.00000223

Explanation:

pKa for C2H5NH3+ = 10.7

pKw = 14.0

pKa + pKb = pKw

10.7 + pKb = 14.0

pKb = 14.0 - 10.7

pKb = 3.30

C2H5NH3Cl is a salt of ethylamine and HCl so it will dissolve in water to produce C2H5NH3^+ + Cl^-

The base hydrolysis reaction: C2H5NH3^+(aq) + H2O(l) <=> C2H5NH2(aq) + H3O^+(aq)

This reaction is described by Kb.

Kb = [C2H5NH2][H3O^+]/[C2H5NH3^+]

Let [C2H5NH2] = [H3O^+] = x,

so [C2H5NH3^+] = 0.26 - x

Kb = x^2/(0.26 - x) = 2.00 x 10^-11

Let's solve for x. In this equation, It is possible to solve without the use quadratic equation. So we can assume that 0.26 - x is approximately equal to 0.26. We won't know until we do the calculation.

We get: x^2 + 2.00 x 10^-11x - 4.99 x 10^-12 = 0

With the use of a quadratic calculator.

x = 2.23 x 10^-6 M = [C2H5NH2] = [H3O^+]

0.26 - x is just 0.26 M in this problem because 2.23 x 10^-6 M is insignificant.

[C2H5NH3^+] = 0.26 M = [Cl^-]

NOTE:

pH = -log [H3O^+] = -log(2.23 x 10^-6) = 5.65

Ka is the acid dissociation constant

Kb is the base dissociation constant

5 0

3 years ago

A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________

gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

Average velocities of gases can be expressed as root-mean-square averages. (V rms)

$\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}$

R = gas constant, T = temperature, Mm = molar mass of the gas particles

From the question

R = 8,314 J / mol K

T = temperature

Mm = molar mass, kg / mol

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

$\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s$

b. the effusion rates of two gases = the square root of the inverse of their molar masses:

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

M₁ = molar mass sulfur dioxide = 64

M₂ = molar mass nitrogen triodide = 395

$\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5$

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0

3 years ago