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Blababa [14]
3 years ago
8

Is the answer A, B, C, or D?

Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
4 0

Answer:

C

Explanation:

C because there is more in it than the rest of the idems

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PLEASE HELP A HOMIIE
Allushta [10]

Answer:

Its the first one. The cells are arranged for structure.

Explanation:

Animal cells do not need to maintain a shape since out bodies do it for us. Plant cells need something for structure.

8 0
3 years ago
Find the volume in milliliters of 2.00 mol of an ideal gas at 36°C and a pressure of 1120 torr.
hram777 [196]

Answer:

V = 34430 mL

Explanation:

Given data:

Volume in mL = ?

Number of moles of gas = 2.00 mol

Temperature = 36°C (36+273= 309K)

Pressure of gas = 1120 torr

Solution:

Formula:

PV = nRT

V = nRT/P

V = 2.00 mol ×62.4 torr • L/mol · K × 309K / 1120 torr

V = 38563.2 torr • L / 1120 torr

V = 34.43 L

L to mL

34.43 L ×1000 mL / 1 L

34430 mL

5 0
3 years ago
In the hospital, 35 babies were born. If 22 were boys, what percentage was boys. Express your answer in places
Hunter-Best [27]
62.857% of the babies were boys
3 0
3 years ago
The indicator propyl red has a Ka of 3.3 X 10-6 . It is red at low pH and yellow at higher pH. What is the approximate pH range
Alik [6]

Answer:

4.48 - 6.48

Explanation:

A pH indicator works in a better way in a range of pH = pKa ± 1. That means we need to determine the pKa of the indicator propyl red to find the range over which it change its color. That is:

pKa = -log Ka

pKa = -log 3.3x10⁻⁶

pKa = 5.48

That means the range over propyl red will change from yellow to red or vice versa is:

4.48 - 6.48

5 0
2 years ago
Determine the vapor pressure (atm) of rubbing alcohol (isopropanol) at 20.0 °C. The normal boiling point of isopropanol is 82.3
jeyben [28]
Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)

The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)

ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
P₂ = 101.2 kPa
8 0
3 years ago
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