Which three elements are combined in lithium sulfate ?

PLEASE HELP

neonofarm [45]

Answer:

The answer to your question is 0.10 M

Explanation:

Data

Molarity = ?

mass of Sucrose = 125 g

volume = 3.5 l

Formula

Molarity = moles / volume

Process

1.- Calculate the molar mass of sucrose

C₁₂H₂₂O₁₁ = (12 x 12) + (1 x 22) + (16 x 11)

= 144 + 22 + 176

= 342 g

2.- Convert the mass of sucrose to moles

342 g of sucrose ------------------- 1 mol

125 g of sucrose -------------------- x

x = (125 x 1) / 342

x = 0.365 moles

3.- Calculate the molarity

Molarity = 0.365 / 3.5

4.- Result

Molarity = 0.10

5 0

3 years ago

How much heat is needed to change 5g of ice at 0°C to water at 50°C?

natta225 [31]

Answer:

400cal

Explanation:

Since ice melted at 0 Celesius then the heat gained by ice (latent heat) will melt it so you should substitute in that law

Q=mlf ..where Q is the heat required to convert ice to water , m is the mass of ice and lf is the latent heat of fusion

Q=5∗80=400cal

6 0

3 years ago

Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of

bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

Step 1: Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

Step 2: The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

Step 3: Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

Step 4: Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

Step 5: Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

Step 6: Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

Step 7: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

Step 8: Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

4 0

3 years ago

X-rays had a frequency of about 3x10^18 determine it's wavelength (in nm) and energy per photon

Serggg [28]

Answer:

E = 19.89×10⁻¹⁶ J

λ = 1×10⁻¹ nm

Explanation:

Given data:

Frequency of xray = 3×10¹⁸ Hz

Wavelength of xray = ?

Energy of xray = ?

Solution:

speed of wave = wavelength × frequency

speed = 3×10⁸ m/s

3×10⁸ m/s = λ ×3×10¹⁸ s⁻¹

λ = 3×10⁸ m/s / 3×10¹⁸ s⁻¹

λ = 1×10⁻¹⁰m

m to nm:

λ = 1×10⁻¹⁰m×10⁹

λ = 1×10⁻¹ nm

Energy of x-ray:

E = h.f

h = plancks constant = 6.63×10⁻³⁴ Js

by putting values,

E = 6.63×10⁻³⁴ Js ×3×10¹⁸ s⁻ ¹

E = 19.89×10⁻¹⁶ J

7 0

2 years ago

Ca(OH)2 (s) precipitates when a 1.0 g sample of CaC2(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g s

Nina [5.8K]

Answer:

D) Ca(OH)₂ will not precipitate because Q < Ksp

Explanation:

Here we have first a chemical reaction in which Ca(OH)₂ is produced:

CaC₂(s) + H₂O ⇒ Ca(OH)₂ + C₂H₂

Ca(OH)₂ is slightly soluble, and depending on its concentration it may precipitate out of solution.

The solubility product constant for Ca(OH)₂ is:

Ca(OH)₂(s) ⇆ Ca²⁺(aq) + 2OH⁻(aq)

Ksp = [Ca²⁺][OH⁻]²

and the reaction quotient Q:

Q = [Ca²⁺][OH⁻]²

So by comparing Q with Ksp we will be able to determine if a precipitate will form.

From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.

mol Ca(OH)₂ = mol CaC₂( reacted = 0.064 g / 64 g/mol = 0.001 mol Ca(OH)₂

the concentration of ions will be:

[Ca²⁺ ] = 0.001 mol / L 0.001 M

[OH⁻] = 2 x 0.001 M = 0.002 M ( From the coefficient 2 in the equilibrium)

Now we can calculate the reaction quotient.

Q= [Ca²⁺][OH⁻]² = 0.001 x (0.002)² = 4.0 x 10⁻⁹

Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸

Therefore no precipitate will form.

The answer that matches is option D

8 0

3 years ago