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Travka [436]
4 years ago
7

Is wine a pure substance

Chemistry
1 answer:
prohojiy [21]4 years ago
6 0

Answer:

yes it is

Explanation:

Wine, air, and gunpowder are other examples of common homogeneous mixtures. Their exact compositions can vary, making them mixtures rather than pure substances. Wine is a liquid mixture of water, ethanol, and a variety of other dissolved substances

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• An element in group 2, period 4 has similar properties with
babymother [125]

Answer:  Elements in Group 2

Explanation:  The periodic table was arranged by Dmitri Mendeleev specifically  around similarites in their chemical behaviors.  He found that as atomic number increases, at some point an element starts to react in a manner similar to a previous one.  When that happened, he would place the larger element under the smaller one, and eventually noticed a periodicity in the table.  Elements in a column (Groups) had similiar chemical properties.  We know today that these similarities are due to the electron configuration, and that these configurations repeat themselves.  He left gaps in the table when he could find an existing element with properties similar to others in that group.  I big leap of faith, but it worked.  Elements for those missing boxes were eventually discovered.

7 0
3 years ago
4 Al + 3O2 → 2Al2O3 If 14.6 grams Al are reacted, how many liters of O2 at STP would be required?
melomori [17]

Answer: 9.08 L

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

\text{Moles of} Al=\frac{14.6g}{27g/mol}=0.54moles

4Al+3O_2\rightarrow 2Al_2O_3

According to stoichiometry :

4 moles of Al require  = 3 moles of O_2

Thus 0.54 moles of Al will require=\frac{3}{4}\times 0.54=0.405moles  of O_2

Standard condition of temperature (STP)  is 273 K and atmospheric pressure is 1 atm respectively.

According to the ideal gas equation:

PV=nRT

P = Pressure of the gas = 1 atm

V= Volume of the gas = ?

T= Temperature of the gas = 273 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 0.405

V=\frac{nRT}{P}=\frac{0.405\times 0.0821\times 273}{1}=9.08L

Thus 9.08 L of O_2 at STP would be required

6 0
3 years ago
How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that
tia_tia [17]

Answer:

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,  

= 2.50 L × 0.800 mol/L

= 2 mol

The given pH of a buffer is 8.53

pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,  

= -log (1.8 ×10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:  

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73= 5.37

[NH₄⁺ = 5.37 × 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,  

mass of NH₄Cl = 10.74 mol × 53.5 g/mol

= 574.59 grams.  

8 0
3 years ago
A halogen is example of a(n)
kenny6666 [7]

A halogen is example of nonmetal. The answer is letter A. When compounds containing halogens they are called salts thus the name “salt – former”. Halogen consists of Fluorine, Chlorine, Bromine, Iodine, Astatine.

6 0
3 years ago
When oxygen and aluminum form an ionic compound, what is the formula? (2 points)
dezoksy [38]
The answer is  D-al2O3
3 0
3 years ago
Read 2 more answers
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