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Oksana_A [137]
3 years ago
8

Formula empirica del glutamato monosódico

Chemistry
1 answer:
Citrus2011 [14]3 years ago
8 0

Answer:

C₅H₈NO₄Na

Explanation:

You might be interested in
the equilibrium concentrations were found to be [H2O]=0.250 M , [H2]=0.600 M , and [O2]=0.800 M . What is the equilibrium consta
antoniya [11.8K]

Answer:

Keq = 0.217

Explanation:

Let's determine the equilibrium reaction.

In gaseous state, water vapor can be decomposed to hydrogen and oxygen and this is a reversible reaction.

2H₂(g)  + O₂(g)  ⇄  2H₂O (g)         Keq

Let's make the expression for the equilibrium constant

Products / Reactants

We elevate the concentrations, to the stoichiometry coefficients.

Keq = [H₂O]² / [O₂] . [H₂]²

Keq = 0.250² / 0.8 . 0.6² =  0.217

3 0
3 years ago
What is the mass of 8.0 x 10^26 UF6 molecules?
gulaghasi [49]
First, find the number of moles of UF6
Avagadro's number = 6.023 x 10^23

Number of moles = 8.0 x 10^26 / Avagadro's number = 8.0 x 10^26 / 6.023 x 10^23 = 1.328 x 10³ moles

Molecular weight of UF6 = Molecular weight of U (238.02891) + Molecular weight of F6 (6 x 18.9984032) = 238.02891 + 113.9904192 = 352.0193292 g/mol

Therefore mass of 8.0 x 10^26 UF6 molecules = 352.0193292 g/mol x 1.328 x 10³ moles = 467.481669 x 10³ grams




3 0
3 years ago
A sample of gas has a volume of 20.0 liters at 22.0° C. If the pressure remains constant, what is the volume at 100.0° C?
miv72 [106K]
The problem applies Charles' law since constant pressure with varying volume and temperature are given. Assuming ideal gas law, the equation to be used is \frac{ V_{1} }{ T_{1} }=\frac{ V_{2} }{ T_{2} }. We make sure the temperatures are expressed in Kelvin, hence the given added with 273. The volume 2 is equal to 25.2881 liters.
6 0
3 years ago
Read 2 more answers
H 2(g)+I 2(g)→2 H I(g)
dlinn [17]

Answer:

The images for the 3 sub-questions have been presented in the first, second and third attached image to this solution.

Explanation:

The reaction for the reaction between Hydrogen gas and Iodine to give Hydrogen iodide

H₂ (g) + I₂ (g) → 2 H I(g)

ΔH∘rxn = −9.5 kJ/mol r x n

a) From the information provided, the heat of reaction being negative shows that the reaction is an exothermic reaction with the products for an exothermic reaction having a lower energy content/level as the reactants for this reaction.

This shows that the energy diagram will have the products at a lower level than fe reactants.

The sketch of the energy diagram for this exothermic reaction is presented in the first attached image to this solution.

The heat of reaction is shown on the energy diagram, together with the activation energy, Ea, the heat content/energy level for the reactants (Hr) & products (Hr) and finally the ΔH∘rxn.

ΔH∘rxn = Hp - Hr

b) When a catalyst that speed up a chemical reaction is introduced to the chemical reaction, the catalyst goes about making the reaction faster by lowering the activation energy of the reaction.

The activation energy is the minimum energy that the reactants must possess to the able to form products. On the graph, it is denoted as Ea, the difference in energy between the peak of the curve and the emergy level of the reactants.

So, a catalyst will cause this activation energy to be lowered to a new level of Ea₁.

The energy diagram of this is presented in the second attached image to this answer.

The ΔH∘rxn remains unchanged for this introduction of catalyst.

The broken lines are used to represent the lowered activation energy action of the catalyst.

c) The reverse reaction will have opposite the energy properties of the initial reaction.

The initially exothermic reaction with higher heat content of reactants than products is now an endothermic reaction with heat content/energy level of the products higher than that of the reactants.

The energy diagram is presented in the third attached image.

The activation energy is bigger and the ΔH∘rxn is now positive and equal to 9.5 kJ/mol, the direct additive inverse of the ΔH∘rxn for the initial exothermic reaction.

Hope this Helps!!!

7 0
3 years ago
How many moles are in 33 L of CO₂ gas at STP?
Leona [35]
We can assume that the volume of one mole of any gas (within reason) is 22.4 L only if we know that the pressure of the gas is 1 atmosphere (101 kPa) and that the temperature of the gas is zero degrees Celsius (273 K). These are known as standard temperature and pressure (STP). So if the gas in question happens to be at STP then the calculation would be:

3.00 L x 1 mol/22.4 L = 0.134 mol

And 3.00 litres CO2 or any other gas at STP would contain 0.134 moles. If the gas was at some other temperature or pressure, assuming they were “moderate,” you could calculate the number of moles by solving the following formula for n.

PV =nRT Where P is pressure, V is volume, T is temperature in Kelvin and R is the ideal gas constant consistent with the units of volume and pressure used.
5 0
3 years ago
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