How many moles are in 564 grams of Copper

If 1 mole of gas occupies 22.4 liters of space, calculate the number of moles that will occupy 11.2 liters of space? Make sure t

Yuliya22 [10]

So, if 1 mole occupies 22.4 L, the imediate conclusion is that a bigger number of moles will occupy more than 22.4 L, and a smaller number of moles will occupy less than 22.4 L. In your case, 3 moles of gas will occupy 3 times more volume than 1 mole of gas

no need to show work i am a teacher at a college

3 0

3 years ago

Aquafers are bodies of water found above ground with thriving ecosystems. O True O False

tekilochka [14]

False.

Aquifers are actually underground! The water from this water-bearing permeable rock is used in wells. Here’s a chart (sorry for the sloppiness).

5 0

3 years ago

The smallest particle of an element that retains the properties of that element is a(n)

alekssr [168]

Answer: atom

Explanation:

It makes the most sense to me anyways seeing that an atom contains protons electrons and neutrons in it already

4 0

3 years ago

2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea

Ivahew [28]

Answer:

The concentration of HI at equilibrium is 0.2445 $\frac{moles}{L}$

Explanation:

A chemical reaction occurs in both directions: from reagents transforming into products (direct reaction) and from products transforming back into reactants (reverse reaction)

The mathematical expression that represents the Chemical Balance is the equilibrium constant Kc.

You have:

aA + bB ⇔ cC + dD

where A, B, C and D represent the chemical species involved and a, b, c and d their respective stoichiometric coefficients. So the constant Kc is:

$Kc=\frac{[A]^{a}*[B]^{b} }{[C]^{c}*[D]^{d} }$

That is, this constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reagents also elevated to their stoichiometric coefficients.

So, in this case:

$Kc=\frac{[H_{2}]*[I_{2} ] }{[HI]^{2} }=0.0156$

You have that the initial concentrations are:

[HI]= $\frac{0.550 moles}{2.00 L} = 0.275 \frac{moles}{L}$

[H₂]= 0

[I₂]= 0

Being "x" the change in the concentration that occurs during the reaction, which must be affected by the stoichiometric coefficient, the final concentrations of each species in equilibrium will be:

[HI]= $0.275 \frac{moles}{L}-x$

[H₂]= x

[I₂]= x

Keep in mind that in the case of reagents the concentration "x" is subtracted because the reagents are consumed. In the case of products, the concentration "x" is added because the reagents are formed.

Then:

$0.0156=\frac{x*x}{(0.275-x)^{2} }$

Resolving

0.0156*(0.275-x)²=x²

0.0156*(0.075625-0.55*x+x²)=x²

1.17975*10⁻³-8.58*10⁻³*x+0.0156*x²=x²

-0.9844*x²-8.58*10⁻³*x+1.17975*10⁻³=0

Solving for x will get you two values: x1≅0.0305 and x2≅-0.0392

Since the value of "x" represents a concentration, and cannot have negative values, the value of x2 is discarded. So: x=x1

Then:

[HI]= $0.275 \frac{moles}{L}-x$ = $0.275 \frac{moles}{L}-0.0305 \frac{moles}{L} = 0.2445 \frac{moles}{L}$

[H₂]= 0.0305 $\frac{moles}{L}$

[I₂]= 0.0305 $\frac{moles}{L}$

The concentration of HI at equilibrium is 0.2445 $\frac{moles}{L}$

4 0

3 years ago

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In the flame test, barium ions produce a green flame whereas calcium ions produce a red flame. In your own words, explain these

deff fn [24]

Answer:

The Barium flame is green because it is a difficult flame to excite, therefore for it to trigger a flame it is necessary that it be too excited for it to occur.

The reddish color of calcium is due to its high volatility and it is sometimes very difficult to differentiate it from strontium.the compression of these elements is due to being able to make them work during combustion

Explanation:

The flame test is a widely used qualitative analysis method to identify the presence of a certain chemical element in a sample. To carry it out you must have a gas burner. Usually a Bunsen burner, since the temperature of the flame is high enough to carry out the experience (a wick burner with an alcohol tank is not useful). The flame temperature of the Bunsen burner must first be adjusted until it is no longer yellowish and has a bluish hue to the body of the flame and a colorless envelope. Then the tip of a clean platinum or nichrome rod (an alloy of nickel and chromium), or failing that of glass, is impregnated with a small amount of the substance to be analyzed and, subsequently, the rod is introduced into the flame, trying to locate the tip in the least colored part of the flame.

The electrons in these will jump to higher levels from the lower levels and immediately (the time that an electron can be in higher levels is of the order of nanoseconds), they will emit energy in all directions in the form of electromagnetic radiation (light) of frequencies characteristics. This is what is called an atomic emission spectrum.

At a macroscopic level, it is observed that the sample, when heated in the flame, will provide a characteristic color to it. For example, if the tip of a rod is impregnated with a drop of Ca2 + solution (the previous notation indicates that it is the calcium ion, that is, the calcium atom that has lost two electrons), the color observed is brick red .

3 0

3 years ago

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