Which of the following characteristics do all plants share?a. being unicellular
b. producing flowers
c. being a prokaryote
d. being an autotroph

Answers

Answer 1

Answer:

The correct option is d. being an autotroph. All plants are multicellular eukaryotes that produce their food through the process of photosynthesis.

What are autotroph ?

Autotrophs are organisms that are capable of producing their own food using energy from the environment, such as sunlight or chemicals. They are also known as "producers" in ecosystems because they are able to synthesize organic molecules from simple inorganic substances.

Plants are one of the most well-known examples of autotrophs, as they use energy from sunlight to produce their own food through photosynthesis.

Not all plants produce flowers, as there are some plant groups that do not produce flowers such as ferns and mosses. Additionally, plants are not unicellular as they are made up of many cells. Finally, plants are not prokaryotes, as they have complex cells with a nucleus and other organelles.

Hence, option d is correct.

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Answer 2

Answer: It would be C because ALL plants are prokaryotes. A is wrong because some plants can be multicellular. B is wrong because not all plants produce flowers. D is wrong because if plants were autotrophs, they would have to eat something to get energy, but plants get energy from the sun.

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How many moles are there in 77.1 g of cl2?

Answers

According to the mole concept, there are 1.8074 moles in 77.1 g of chlorine.

What is a mole?

Mole is defined as the unit of amount of substance . It is the quantity measure of amount of substance of how many elementary particles are present in a given substance.

It is defined as exactly 6.022×10²³ elementary entities. The elementary entity can be a molecule, atom ion depending on the type of substance. Amount of elementary entities in a mole is called as Avogadro's number.

It is widely used in chemistry as a suitable way for expressing amounts of reactants and products.For the practical purposes, mass of one mole of compound in grams is approximately equal to mass of one molecule of compound measured in Daltons. Number of moles is calculated as, mass/molar mass

Substituting values in formula gives, number of moles=77.1/70.90=1.8074

Thus, there are 1.8074 moles in 77.1 g of Cl₂.

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There are 70.90g of Cl2 in 1 mol (because 35.45g Cl in one mol). So 77.1g/70.9g≈1.0874 mol.

A gas that extinguishes a flame is hydrogen. true or false

Answers

This is true because it reacts with oxygen

What was a direct result of the Bay of Pigs invasionin 1961?
(1) Fidel Castro was removed from power.
(2) Cold War tensions increased.
(3) The United States announced its Good
Neighbor policy.
(4) The communist government in Nicaragua was
overthrown.

Answers

A direct result of the Bay of Pigs invasion in 1961 was that "Cold War tensions increased", mainly because this put into question the boundaries of both states.

2) Cold War tensions increased.

The Bay of Pigs invasion was carried out by Cuban exiles that had been trained by the United States' Central Intelligence Agency (CIA). The Cuban counter revolutionaries failed and Kennedy was embarrassed. Within just days, the force of exiles was soundly defeated by the Cuban Revolutionary Armed Forces, commanded by Fidel Castro. In the aftermath of the Bay of Pigs affair, Cuba requested that the USSR, as its communist sponsor, place missiles in Cuba to deter further US-sponsored action against the Castro regime.

2KI + Pb(NO3)2 → 2KNO3 + PbI2 Determine how many grams of KI are required to produce 1.2g of PbI2.

Answers

Molar mass KI = 166.0028 g/mol and Pbl₂ = 461.0089 g/mol

2 Kl + Pb(NO₃)₂ ------> 2 KNO₃ + Pbl₂

2 x 166.0028 g ---------> 461.0089 g
? g KI -------------> 1.2 g

Mass KI = 1.2 x 2 x 166.0028 / 461.0089

Mass KI = 398.40672 / 461.0089

Mass KI = 0.86 g

hope this helps!

Which mineral reacts to hydrochloric acid (HCl) only when powdered?

Answers

Answer:The mineral that reacts to hydrochloric acid (HCl) only when powdered is calcite. Calcite is a carbonate mineral composed of calcium carbonate (CaCO3). When calcite is in its powdered form, it readily reacts with hydrochloric acid to produce carbon dioxide (CO2), water (H2O), and calcium chloride (CaCl2). This reaction can be described by the chemical equation:

CaCO3 (s) + 2HCl (aq) → CO2 (g) + H2O (l) + CaCl2 (aq)

The reaction occurs because the surface area of the powdered calcite is increased, allowing for a greater contact area between the mineral and the hydrochloric acid. This increased contact area facilitates a faster and more vigorous reaction compared to when the calcite is in its solid, non-powdered form.

It's important to note that not all minerals react with hydrochloric acid. Only minerals that contain carbonate ions (CO3^2-) will react with hydrochloric acid to produce carbon dioxide gas. Other common minerals that exhibit this reaction include limestone and marble, which also contain calcium carbonate.

By understanding this reaction and its characteristics, you can identify calcite and other carbonate-containing minerals by their reaction to hydrochloric acid when in powdered form.

Final answer:

Calcite is a mineral that reacts to hydrochloric acid (HCl) only when powdered. This reaction produces carbon dioxide gas.

Explanation:

The mineral that reacts to hydrochloric acid (HCl) only when powdered is calcite.

When solid calcite is exposed to hydrochloric acid, it does not undergo any noticeable reaction. However, when powdered calcite is mixed with HCl, it readily fizzes and releases carbon dioxide gas.

This reaction occurs because the acid dissolves the calcite, converting it into dissolved calcium ions and carbon dioxide gas.

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An element crystallizes in a face centered cubic lattice and has a density of 1.45 g cm-3 . The edge of its unit cell is 4.52 x 10-8cm.a) How many atoms are in each unit cell?
b) What is the volume of a unit cell?
c) What is the mass of a unit cell?
d) Calculate the approximate atomic mass of the element.

Answers

In the given fcc element, a. the number of atoms is 4. b. The volume of a unit cell is $\rm 9.23\;*\;10^-^2^3\;cm^3$ . c. Mass of unit cell is $\rm 1.34\;*\;10^-^2^2\;g$ . d. The approximate atomic mass of the element is 80.7 amu.

The face-centered cubic lattice has 3 atoms from the 6 faces, and 1 atom from the eight corners. Thus, the total atoms in the face-centered lattice are four.

The face-centered lattice has been a cube.

The volume of cube = $\rm (edge)^3$

The volume of unit cell = $\rm (4.52\;*\;10^-^8\;cm)$

The volume of unit cell = $\rm 9.23\;*\;10^-^2^3\;cm^3$

The mass of a unit cell can be calculated from density. Mass can be defined as the ratio of volume to density.

Mass = $\rm (volume)/(density)$

Mass of unit cell = $\rm (9.23\;*\;10^-^2^3\;cm^3)/(1.45\;g\;cm^-^3)$

Mass of unit cell = $\rm 1.34\;*\;10^-^2^2\;g$ .

The approximate atomic mass of the element can be calculated by the mass of the carbon atom.

Mass of 1 carbon atom = $\rm (mass\;of\;1\;mole\;carbon)/(number\;of\;atoms\;in\;1\;mole\;Carbon)$

Mass of 1 carbon atom = $\rm (12)/(6.023\;*\;10^2^3)$

Mass of 1 carbon atom = 1.992 $\rm *\;10^-^2^3$ grams.

atomic mass unit per gram can be given as;

amu/gram = $\rm (12)/(1.992\;*\;10^-^2^3)$

amu/gram = $\rm 6.022\;*\;10^2^3$ amu/gram

1 gram = $\rm 6.022\;*\;10^2^3$ amu

1 amu = 1.661 $\rm *\;10^-^2^4$ gram.

The average atomic mass = mass of unit cell $*$ amu\gram

= $\rm 1.34\;*\;10^-^2^2\;g$ . $*$ 1 amu/ 1.661 $\rm *\;10^-^2^4$ gram.

= 80.7 amu.

In the given fcc element, a. the number of atoms is 4. b. The volume of a unit cell is $\rm 9.23\;*\;10^-^2^3\;cm^3$ . c. Mass of unit cell is $\rm 1.34\;*\;10^-^2^2\;g$ . d. The approximate atomic mass of the element is 80.7 amu.

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Final answer:

A face centered cubic lattice consists of 4 atoms. The volume of the unit cell is 9.22 x 10^-23 cm^3 and the mass is 1.34 x 10^-23 g. The approximate atomic mass of the element is 2.02 amu.

Explanation:

The element is said to crystallize in a face centered cubic lattice. This implies that there is one atom at each corner of the cube (8 corners for a total of 1 atom, since each corner atom is shared among 8 adjacent cubes). There is also one atom on each face of the cube (6 faces for a total of 3 atoms, since each face atom is shared among 2 adjacent cubes). Thus, a total of 4 atoms are present in each unit cell. (a)

The volume of a unit cell (edges for a cube) can be calculated by the formula 'volume = side^3', where side in this case is 4.52 x 10-8cm. Hence, the volume equals (4.52 x 10^-8cm)^3 = 9.22 x 10^-23cm^3. (b)

The density of the substance is given as 1.45g/cm^3. The formula for density is 'mass/volume' which implies that mass can be calculated as 'density x volume'. Hence, the mass of the unit cell is (1.45g/cm^3) x (9.22 x 10^-23 cm^3) = 1.34 x 10^-23 g.(c)

The atomic weight of the element can then be calculated by taking this overall mass and dividing by the number of atoms in a unit cell (4). So, the atomic weight is (1.34 x 10^-23 g) / 4 = 3.35 x 10^-24 g. But atomic weights are usually given in atomic mass units (amu), not grams, and 1 amu = 1.66 x 10^-24 g. Therefore, we have an atomic weight of (3.35 x 10^-24 g) / (1.66 x 10^-24 g/amu) = approximately 2.02 amu. (d)