In the hydrogenation of oils the catalyst used is: A. Pt B. Ni C. Fe D. V2O5

What is the molar mass of SO4?

VLD [36.1K]

Yo sup??

the answer is option A ie

96.056 grams/mole

because mass of S is 32 gm and mass of O is 16 gm

Hope this helps

3 0

3 years ago

And can The law of conservation of mass states that neither be nor

Harlamova29_29 [7]

Answer:

The law of conservation of mass states that the quantity of the mass can neither be added to, neither can it be removed. The greater picture's mass will always be the same amount.

6 0

3 years ago

When water is heated in an ocean the liquid water changes from and turns into?

sasho [114]

Hello!

The answer is Evaporation

When you add heat to water, it boils, releasing gas into the air. This is called Evaporation. The water changes from a liquid to a gas.

Hope this helped!

6 0

3 years ago

A student mixes in a test tube 3.00mL of 0.050M CuSO4with 7.00mL of 0.20M NH3/NH41 . The solution becomes a deep blue color. Ass

valkas [14]

Answer:

$\large \boxed{\text{0.0035 mol/L}}$

Explanation:

We are given the volumes and concentrations of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

Cu²⁺ + 4NH₃ ⟶ Cu(NH₃)₄²⁺

V/mL: 3.00 7.00

c/mol·L⁻¹: 0.050 0.20

1. Identify the limiting reactant

(a) Calculate the moles of each reactant

$\text{Moles of Cu}^{2+}= \text{3.00 mL solution} \times \dfrac{\text{0.050 mmol Cu}^{2+}}{\text{1 mL solution}} = \text{0.150 mmol Cu}^{2+}\\\\\text{Moles of NH}_{3} = \text{7.00 mL solution} \times \dfrac{\text{0.20 mmol NH}_{3}}{\text{1 mL solution}} = \text{0.140 mmol NH}_{3}$

(b) Calculate the moles of Cu(NH₃)₄²⁺ that can be formed from each reactant

(i) From Cu²⁺

$\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.150 mmol Cu}^{2+} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{1 mmol Cu}^{2+}}\\\\= \text{0.150 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}$

(ii) From NH₃

$\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.140 mmol NH}_{3} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{4 mmol NH}_{3}}\\\\= \text{0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}$

NH₃ is the limiting reactant, because it forms fewer moles of the complex ion.

(c) Concentration of the complex ion

$\text{The reaction forms 0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$ in a total volume of 10.00 mL.}\\c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.0350 mmol}}{\text{10.00 mL}} = \textbf{0.0035 mol/L}\\\\\text{The concentration of the complex ion is $\large \boxed{\textbf{0.0035 mol/L}}$}$

7 0

3 years ago

Which of the following occurs when a reaction in a solution is at equilibrium and more product is added to the solution?No chang

LenaWriter [7]

Answer:

Equilibrium shifts to produce more reactant

Explanation:

Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

When more product is added to the solution:

This will increase the concentration of the products side, so the reaction will be shifted to the lift side (reactants side) to suppress the increase in the concentration of Products.

So, the right choice is: Equilibrium shifts to produce more reactant

3 0

3 years ago

In the hydrogenation of oils the catalyst used is: A. PtB. Ni C. FeD. V2O5​

In the hydrogenation of oils the catalyst used is: A. PtB. Ni C. FeD. V2O5