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docker41 [41]
4 years ago
9

Flowering plants are the most successful group of land plants. This is due, in part, to the trait that could be the label for po

int C. in the cladogram. Consider the organization of the cladogram. What trait, besides flowers, is unique to this group of plants?
Chemistry
1 answer:
jek_recluse [69]4 years ago
8 0

Answer:

seeds are protected in fruits in such kind of plants.

Explanation:

The flowering plants are also known as Angiospermae, or Magnoliophyta. Flowering plants are plants that grow flowers. Such plants use seeds to reproduce, or make more plants like them.

Seeds are protected in fruits in such kinds of plants.

Examples of flowering plants are daisies, tulips, oaks, apples.

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What is the ideal gas law?
olga nikolaevna [1]

Answer:

Option D. A law describing the behavior of condensed gases

Explanation:

Ideal gas law describes the behavior of gas molecules when the pressure is applied on it.

Ideal gas law states that the product of the volume and the pressure of molecule of an ideal gas is equal to the product of the temperature of the gas and the universal gas constant.

The equation of ideal gas law is given below.

PV = nRT

3 0
3 years ago
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Which property describes the way substances react with other substances to form new substances?
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Answer:

Chemical Property describes the way substances react with other substances to form new substances.

Explanation:

Hope it helps you

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3 years ago
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The freezing point of water H2O is 0.00°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in water is urea. If 13.
Trava [24]

Answer:

Molality = 1.46 molal

The freezing point of the solution = -2.72 °C

Explanation:

Step 1: Data given

The freezing point of water H2O is 0.00°C at 1 atmosphere

urea = nonelectrolyte = van't Hoff factor = 1

Mass urea = 13.40 grams

Molar mass urea = 60.1 g/mol

Mass of water = 153.2 grams

Molar mass H2O = 18.02 g/mol

Kf = 1.86 °C/m

Step 2: Calculate moles urea

Moles urea = mass urea /molar mass urea

Moles urea = 13.40 grams / 60.1 g/mol

Moles urea = 0.223 moles

Step 3: Calculate the molality

Molality = moles urea / mass water

Molality = 0.223 moles / 0.1532 kg

Molality = 1.46 molal

Step 4: Calculate the freezing point of the solution

ΔT = i * Kf * m

ΔT = 1* 1.86 °C/m * 1.46 m

ΔT = 2.72 °C

The freezing point = -2.72 °C

3 0
3 years ago
A sample of a compound contains 9.11 g Ni and 5.89 g F. What is the empirical formula of this compound?
Novay_Z [31]
NiF2 is the correct answer for the is problem
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3 years ago
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A biochemist who is studying the properties of certain sulfur (s)-containing compounds in the body, wonders whether trace amount
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Elements of the same group show similar chemical properties as they have the same number of valence electrons. Sulfur belongs to the group 16 or 6 A or the oxygen family with 6 valence electrons. The elements of group 6 A are Oxygen, Sulfur, Selenium, Tellurium and Polonium. All of these elements show similar chemical properties. Therefore, when a biochemist studying the properties a sulfur containing biochemical compounds in the body wants to look at any other non-metal with similar properties, he has to consider other elements of the group 6 A like Oxygen (O), Selenium (Se), Tellurium (Te) and Polonium (Po).

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3 years ago
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