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DENIUS
1 day ago
6

Consider butter (density= 0.860 g/mL) and sand (density= 2.28 g/mL). If 1.00 mL of butter were mixed with 1.00 mL of sand and mi

xed as thoroughly as possible what would be the density of the resulting mixture?
Chemistry
1 answer:
Anarel [852]1 day ago
6 0

The mixture’s density is 1.57 g/cm³.


Step 1: Determine the mass of the butter.


\text{Mass} = \text{1.00 cm}^{3 } \times \frac{\text{0.680 g} }{\text{1 cm}^{3 }} = \text{0.860 g}\\

Step 2: Determine the mass of the sand.


\text{Mass} = \text{1.00 cm}^{3 } \times \frac{\text{2.28 g} }{\text{1 cm}^{3 }} = \text{2.28 g}\\

Step 3: Determine the density of the mixture.

Total mass = 0.860 g + 2.28 g = 3.14 g.

Total volume = 1 cm³ + 1 cm³ = 2 cm³

\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{\text{3.14 g} }{\text{2 cm}^{3 }} = \textbf{1.57 g/cm}{^{3}\\

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A certain alcoholic beverage contains only ethanol (C2H6O) and water. When a sample of this beverage undergoes combustion, the e
castortr0y [923]

Response:

9.606 g

Clarification:

Step 1: Write the balanced combustion equation

C₂H₆O(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(g)

Step 2: Determine the moles for 11.27 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

11.27 g × (1 mol/18.02 g) = 0.6254 mol

Step 3: Find the moles of C₂H₆O that produced 0.6254 moles of H₂O

The ratio of C₂H₆O to H₂O is 1:3. Thus, the moles of C₂H₆O are 1/3 × 0.6254 mol = 0.2085 mol

Step 4: Calculate the mass for 0.2085 moles of C₂H₆O

The molar mass of C₂H₆O is 46.07 g/mol.

0.2085 mol × 46.07 g/mol = 9.606 g

7 0
10 days ago
Consider the following system at equilibrium:
VMariaS [1037]

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and cut (q) down to a third

Explanation:

According to the principle of Le Chatelier, when a system reaches equilibrium and a change is introduced, the system will respond to counteract that change.

Since P and Q are reactants, raising the amount of either one or both without a proportional rise in R (which is a product) will cause the equilibrium to move towards the right. Similarly, if R decreases while P and Q remain constant, this too will push the equilibrium to the right. Thus, Increase(P), Increase(q), and Decrease(R) will lead to a rightward shift in the equilibrium.

Conversely, raising R without increasing P and Q will draw the equilibrium to the left. Likewise, cutting down P and/or Q without a similar reduction in R will shift the equilibrium leftward. Therefore, Increase(R), Decrease(P), Decrease(q), and triple both (Q) and (R) will shift the equilibrium to the left.

If there are equivalent changes in P and Q, with R remaining unchanged, then the equilibrium remains stationary. So, tripling (P) while reducing (q) to one third will not alter the equilibrium.

6 0
3 days ago
Hitesh wants to learn swimming. He brought swimming costumes and a cap. What kind of material should these be made of and why?
alisha [964]

Answer:

Nylon and Spandex (Lycra).

Explanation:

These materials are designed to fit the body, with nylon drying more quickly than other types of fabrics, and Spandex being commonly found in swimming and sports apparel due to its elastic qualities. Both fabrics also wick moisture away and dry rapidly.

With high capacity and enhanced flexibility, nylon and Spandex provide a snug fit to the body and can retain their shape during various activities, making them ideal for swimming.

This explains why these materials are suitable based on the situation given.

3 0
13 days ago
The next 3 questions will walk you through using the Henderson-Hasselbalch equation for the following question. For each step pr
lions [985]

Response:

The pKa value is 13.0.

Clarification:

pKa + pKb = 14

For trimethylamine, Kb = 6.3 × 10^{-5}

Calculating pKb: pKb = - log (6.3 × 10^{-5})

= 1.0

Thus, pKa = 14 - pKb = 14 - 1.0

pKa = 13.0

Verification: The typical range for pKa in weak acids is from 2 to 13.

8 0
5 days ago
A gas sample of argon, maintained at constant temperature, occupies a volume of 500. l at 4.00 atm. what is the new volume if th
lorasvet [956]
Boyle's law describes the relationship between gas pressure and volume. 
It asserts that at a constant temperature, pressure is inversely proportional to gas volume.
PV = k
where P represents pressure, V denotes volume, and k is a constant. 
P1V1 = P2V2
where the parameters for the initial condition are on the left, and the parameters for the second condition appear on the right side of the formula.
By substituting values into the equation: 4.00 atm x 500 L = 8.0 atm x V
V calculates to 250 L.
Thus, the new volume becomes 250 L.
6 0
6 days ago
Read 2 more answers
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