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Gekata [30.6K]
3 years ago
9

Many plants that grow in shady areas have large leaves. This feature helps the plants gather more sunlight for photosynthesis. W

hich statement describes an example of evolution?
A. All plants pass the genes for large leaves to their offspring.
B. Plants with leaves of varying size sprout in gardens each year.
C. All plants have variation in their genes for leaf size.
D. Plants with large leaves have become common in rain forests.
Chemistry
1 answer:
Ilya [14]3 years ago
8 0

Answer: D.Plants with large leaves have become common in rain forest

Explanation: I got this right on the test trust me

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To act as a battery, a redox reaction must have a negative voltage.<br> True or False?
LenaWriter [7]

Answer:

False

Explanation:

6 0
3 years ago
1) What is the mass of 6.2 mol of K2CO3?
lorasvet [3.4K]

Answer:

\boxed {\boxed {\sf 856.8648 \ grams \ of \ K_2CO_3}}

Explanation:

To convert from moles to grams, we must find the molar mass.

1. Molar Mass

First, identify the elements in the compound. K₂CO₃ It has potassium, carbon, and oxygen. Find these elements and their masses on the Periodic Table.

  • K: 39.098 g/mol
  • C: 12.011 g/mol
  • O: 15.999 g/mol

Note the subscript of 2 after K and 3 after O. We must multiply oxygen's molar mass by 2, then oxygen's by 3, and add carbon.

  • 2(39.098 g/mol) + 3(15.999 g/mol) + 12.011 g/mol= 138.204 g/mol

2. Convert Moles to Grams

Use the molar mass as a fraction.

\frac {138.204 \ g \ K_2CO_3}{1 \ mol \ K_23CO_3}

Multiply by the given number of moles: 6.2

6.2 \ mol \ K_2CO_3 *\frac {138.204 \ g \ K_2CO_3}{1 \ mol \ K_23CO_3}

6.2  *\frac {138.204 \ g \ K_2CO_3}{1 }

6.2  * {138.204 \ g \ K_2CO_3}

856.8648 \ g \ K_2CO_3

There are <u>856.8648 grams</u> of potassium carbonate in 6.2 moles.

4 0
3 years ago
What are the chemical properties of beryllium phosphide??
o-na [289]

Answer:

Forms ionic bonds, insoluble, doesn't participate in single or double displacement reactions, non-reactive, high heat of combustion

Explanation:

Beryllium is a metal, since it belongs to group 2A, the alkaline earth metals. It has a total of 2 valence electrons.

Phosphorus, on the other hand, belongs to group 5A and has a total of 5 valence electrons.

We have a compound which has a metal in it, therefore, it's an ionic compound. Beryllium, our metal, loses its 2 electrons to gain an octet and phosphorus, our nonmetal, should gain 3 electrons to have an octet. The oxidation states are +2 and -3 respectively. This means we need 3 beryllium cations and 2 phosphide anions in our formula Be_3P_2.

Beryllium phosphide would be expected to be insoluble, as only beryllium chloride, fluoride, nitrate, phosphate and sulfate are soluble substances, while the remaining ones are expected to be insoluble.

Due to its insolubility, beryllium phosphide would not participate in any ionic reactions, such as single displacement or double displacement.

Since it's insoluble, we expect this compound to be chemically stable and not reactive. This implies that if we wanted to burn it, the heat of combustion would be very high, as a lot of energy would be needed to be supplied in an endothermic reaction in order to burn it.

8 0
3 years ago
The crystal size of an igneous rock is described as its _____.​
Mila [183]

Answer:

The crystal size of an igneous rock is described as its texture.

4 0
3 years ago
Read 2 more answers
What is the new freezing point for a 0.811m solution of Li2O in water?
rewona [7]

Answer:

Freezing T° of solution = - 4.52°C

Explanation:

ΔT = Kf . m . i

That's the formula for colligative property about freezing point depression.

Li₂O is an oxide that can not be dissociated but, if we see it's a ionic compound.

Li₂O →  2Li⁺  +  O⁻²

3 moles of ions have been formed. Ions dissolved in solution are i, what we call Van't Hoff factor.

m is molality → 0.811 m, this is data

Kf →Cryoscopic constant, for water is 1.86 °C/m

and ΔT = Freezing T° of pure solvent - Freezing T° of solution

We replace: 0°C - Freezing T° of solution = 1.86°C/m . 0.811 m . 3

Freezing T° of solution = - 4.52°C

3 0
3 years ago
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