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3 years ago

Which of these is a physical property of a tree?

Chemistry

2 answers:

Anit [1.1K]3 years ago

5 0

Is there answer choices?

Nezavi [6.7K]3 years ago

4 0

Nothing is here bud. But here are some answers anyways.

leaves (some trees)
branches
roots
sap (some trees)
Brown bark

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23. SIO2 What is the compound?

xxMikexx [17]

Silicon dioxide is the compound

6 0

3 years ago

Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.3 g of methane is

Marrrta [24]

Answer:

26 g

Explanation:

Write the balanced reaction first

CH4 + 2 O2 --> CO2 + 2 H2O

9.3g + 52.3g --> ? CO2

You must determine how much carbon dioxide can be made from each of the reactants. The maximum mass that can be made is the lower of the two.

From CH4:

9.3g CH4 (1molCH4/16.05gCH4) (1molCO2 / 1molCH4) (44.01g CH4 / 1molCO2) = 26 g

From O2:

52.3g O2 (1molO2/32gO2) (1molCO2/2molO2)(44.01g/1molCO2) = 36 g

6 0

3 years ago

A sample of glucose ( C6H12O6 ) of mass 8.44 grams is dissolved in 2.11 kg water. What is the freezing point of this solution? T

taurus [48]

Answer:

- 0.0413°C ≅ - 0.041°C (nearest thousands).

Explanation:

Adding solute to water causes the depression of the freezing point.

We have the relation:

ΔTf = Kf.m,

Where,

ΔTf is the change in the freezing point.

Kf is the freezing point depression constant (Kf = 1.86 °C/m).

m is the molality of the solution.

Molality is the no. of moles of solute per kg of the solution.

no. of moles of solute (glucose) = mass/molar mass = (8.44 g)/(180.156 g/mol) = 0.04685 mol.

∴ molality (m) = no. of moles of solute/kg of solvent = (0.04685 mol)/(2.11 kg) = 0.0222 m.

∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.

∴ The freezing point of the solution = the freezing point of water - ΔTf = 0.0°C - 0.0413°C = - 0.0413°C ≅ - 0.041°C (nearest thousands).

6 0

3 years ago

A student used 1.168g of an unknown weak acid and titrated it against 28.75 mL of 0.105M NaOH to reach the equivalence point. Wh

Vaselesa [24]

Answer:

The molar mass of the unknown acid is 386.8 g/mol

Explanation:

Step 1: Data given

Mass of the weak acid = 1.168 grams

volume of NaOH = 28.75 mL = 0.02875 L

Molarity of NaOH = 0.105 M

Since we only know 1 equivalence point, we suppose the acid is monoprotic

Step 2: Calculate moles NaOH

Moles NaOH = molarity NaOH * volume NaOH

Moles NaOH = 0.105 M * 0.02875 L

Moles NaOH = 0.00302 moles

We need 0.00302 moles of weak acid to neutralize the NaOH

Step 3: Calculate molar mass of weak acid

Molar mass = mass / moles

Molar mass = 1.168 grams / 0.00302 moles

Molar mass = 386.8 g/mol

The molar mass of the unknown acid is 386.8 g/mol

3 0

3 years ago

Ice cream is made by freezing a liquid mixture that, as a first approximation, can be considered a solution of sucrose in water.

hjlf

Answer:

Freezing point is -2.81°C

Explanation:

34g/342gmol^-1 = 0.0994mol

n = m/mr

Molarity= 0.994/ 0.66 = 1.51M

◇T = -i × m ×Kf

Where ◇T is freezing depression

i= Vant Hoff factor

m = molarity

Kf = freezing content = 1.

860kgmol^-1

◇T =-1 × 1.51 × 1.860 = - 2.81°C

6 0

3 years ago