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harkovskaia [24]
2 years ago
6

If 4.59 g of potassium reacts with 3.6 g of sulfur according to the following reaction, how many grams of potassium sulfide can

theoretically be produced? Please identify the limiting and excess reactants: 2K + S -> K^2S
Chemistry
1 answer:
EleoNora [17]2 years ago
8 0
First convert the amount of grams you have of each substance to moles. Find your limiting reactant by calculating how many grams are needed to complete this reaction. If done correctly, you would see that we need .226 moles of Potassium to complete this reaction. However, we only have .118 moles of Potassium, so K must be our limiting reactant. Then use the moles of K to find out how many moles of K^2S are made. Then convert the amount of moles of K^2S to grams and you should get 10.3 g K^2S
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If a cell is isotonic with a 0.88% nacl solution, how would an extracellular fluid with 1% nacl affect the cell?
EleoNora [17]
<span>The extracellular fluid is high in NaCl so the cell would be dehydrated further and the two solutions would equilibrate. Ultimately water would leave the cell and passes to </span>extracellular fluid and equilibrium is reached.
3 0
2 years ago
Read 2 more answers
6. The half life for uranium-235 is 7.0x108 years.
kaheart [24]

Answer:

\large \boxed{4.0}

Explanation:

\text{No. of half-lives} = 2.8  \times 10^{9 } \text{ yr} \times \dfrac{\text{1 half-life}}{7.0 \times 10^{8} \text{ yr}} = \textbf{4.0 half-lives}\\\\\text{The sample went through $\large \boxed{\textbf{4.0 half-lives}}$}

4 0
2 years ago
Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying t
-Dominant- [34]

Answer:

The description including its given problem is outlined in the following section on the clarification.

Explanation:

The given values are:

RBCC = 0.12584 nm

RFCC = 0.12894 nm

The unit cell edge length (ABCC) as well as the atomic radius (RBcc) respectively connected as measures for BCC (α-phase) structure:

√3 ABCC = 4RBCC

⇒  ABCC = \frac{4RBCC}{\sqrt{3} }

⇒             = \frac{4\times 0.12584}{\sqrt{3}}

⇒             = 0.29062 \ nm

Likewise AFCC as well as RFCC are interconnected by  

√2AFCC = 4RFCC

⇒  AFCC = \frac{4RFCC}{\sqrt{2}}

⇒             = \frac{4\times 0.12894}{\sqrt{2} }

⇒             = 0.36470 \ nm

Now,

The Change in Percent Volume,

= \frac{V \ final-V \ initial}{V \ initial}\times 100 \ percent

= \frac{(VFCC)unit \ cell-(VBCC)unit \ cell}{(VBCC)unit \ cell}\times 100 \ percent

= \frac{(aFCC)^3-(aBCC)^3}{(aBCC)^3}\times 100 \ percent

= \frac{(0.36470)^3-(0.29062)^3}{(0.29062)^3}\times 100 \ percent

= 97.62 \ percent (approximately)

Note: percent = %

7 0
2 years ago
4.8g of calcium is added to 3.6g of water. The following reaction occurs
notka56 [123]
Q1)
the number of moles can be calculated as follows
number of moles = mass present / molar mass
number of moles is the amount of substance.
4.8 g of Ca was added therefore mass present of Ca is 4.8 g
molar mass of Ca is 40 g/mol 
molar mass is the mass of 1 mol of Ca
therefore if we substitute these values in the equation 
number of moles of Ca = 4.8 g / 40 g/mol = 0.12 mol
0.12 mol of Ca is present 

q2)
next we are asked to calculate the number of moles of water present 
again we can use the same equation to find the number of moles of water
number of moles = mass present / molar mass
3.6 g of water is present 

sum of the products of the molar masses of the individual elements by the number of atoms 
H - 1 g/mol and O - 16 g/mol 
molar mass of water = (1 g/mol x 2 ) + 16 g/mol = 18 g/mol 
molar mass of H₂O is 18 g/mol 
therefore number of moles of water  = 3.6 g / 18 g/mol = 0.2 mol 
0.2 mol of water is present 
8 0
2 years ago
Karen has discovered a new organism in the Amazon Rainforest. It moves by itself, is a consumer, and its cells do not have a cel
EastWind [94]

The correct answer is option B, that is, Animalia.  

The animals refer to the multicellular eukaryotic species, which forms the biological kingdom known as Animalia. The majority of the animals breathe oxygen, consume organic substances, and possess the tendency to move, has the tendency to reproduce, and develop from a hollow sphere of cells, the blastula at the time of embryonic development. In comparison to Kingdom Plantae, the cells in animals do not exhibit cell wall. Thus, on the basis of the characteristics, the newly discovered species must be categorized under Kingdom Animalia.  


3 0
2 years ago
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