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2 years ago

_____________ is a common and tasty compound made up of the elements carbon, hydrogen, and oxygen

Chemistry

1 answer:

ziro4ka [17]2 years ago

8 0

Answer:

Glucose

Explanation:

Glucose, also called dextrose is one of a group of carbohydrates known as simple sugars or monosaccharides. Glucose is derived from the Greek word 'glykys' meaning “sweet”. It has the molecular formula C6H12O6. It is commonly found in fruits and honey and is the major free sugar circulating in the blood of higher animals. It is the source of energy in cell function, and the regulation of its metabolism is of great importance. Molecules of starch, the major energy-reserve of carbohydrate of plants, consist of thousands of linear glucose units. Another major compound composed of glucose is cellulose, which is also linear. Dextrose is the molecule D-glucose.

Glucose is composed of carbon hydrogen and oxygen in a ratio of 1:2:1 as typified by its formula. It is known for its sweet taste. It can be combined with fructose in table sugar.

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Filtration, or screening, is one way to separate certain mixtures, as shown in the image. Which of these mixtures could be most

erastovalidia [21]

Answer: D

Explanation:

3 0

3 years ago

Column A 1. Reproduction: Reproduction 2. offspring: offspring 3. gamete: gamete 4. Budding: Budding 5. Binary fission: Binary f

7nadin3 [17]

Answer:

1. Reproduction: One or two parents, organisms make new offspring

2. Offspring: A new living organism

3. Gamete: female and male create mini-me by cell division

4. Budding: Hydra

5. Binary fission: Bacteria

6. Zygote: The original cell after egg cell and sperm cell join

7. Fragmentation/Regeneration: Sea star / Planarian

8. Mitosis: Healling and growth occurs in budding

9. Meiosis: Egg or sperm cell/ specialized cells

Explanation:

1. Reproduction is the process by which organisms make new offspring either sexually (involving two parents) or asexually (involving one parent only).

2. An Offspring is a new living organism

3. A gamete is a mature sexual reproductive cell, as a sperm or egg, that unites with another cell to form a new organism. It formed from female and male parents by cell division known as meiosis

4. Budding is a form of asexual reproduction in which a new offspring grows out from a part of the parent. It occurs in Hydra.

5. Biinary fission is a form of asexual reproduction which involves a separation of the parent into two new offspring. It commonly occurs in Bacteria

6. Zygote is the original cell formed after egg cell and sperm cell fuse.

7. Fragmentation/Regeneration: this is a form of reproduction in which a parent splits into fragments which are then able to develop into new organisms. It occurs in Sea star and Planarians

8. Mitosis is a form of cell division which a cell divides into two identical daughter cells with the same genetic components as the parent cell. It functions in healing and growth as well as in budding

9. Meiosis is a type of cell division that results in the production of sex cells or gametes. The number of chromosomes in the parent cell is halved and four gamete cells are produced. Egg or sperm cell/ specialized cells are produced by meiosis.

5 0

3 years ago

What is the molarity of a solution containing 55.8 g of mgcl2 dissolved in 1.00 l of solution?

MakcuM [25]

The answer is 0.59 M.

Molar mass (Mr) of MgCl₂ is the sum of the molar masses of its elements.
So, from the periodic table:
Mr(Mg) = 24.3 g/l
Mr(Cl) = 35.45 g/l
Mr(MgCl₂) = Mr(Mg) + 2Mr(Cl) = 24.3 + 2 · 35.45 = 24.3 + 70.9 = 95.2 g/l

So, 1 mol has 95.2 g/l.

Our solution contains 55.8g in 1 l of solution, which is 55.8 g/l

Now, we need to make a proportion:
1 mole has 95.2 g/l, how much moles will have 55.8 g/l:
1 M : 95.2 g/l = x : 55.8 g/l
x = 1 M · 55.8 g/l ÷ 95.2 g/l ≈ 0.59 M

7 0

3 years ago

Read 2 more answers

A sample of gas has a volume of 20.0 mL at STP. What will the volume be if the temperature is changed to 546 K and the pressure

Ostrovityanka [42]

The volume did not change, it remained at 20 ml

<h3>Further explanation</h3>

Given

20 ml a sample gas at STP(273 K, 1 atm)

T₂=546 K

P₂=2 atm

Required

The volume

Solution

Combined gas Law :

$\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}$

Input the value :

$\tt \dfrac{1\times 20}{273}=\dfrac{2\times V_2}{546}\\\\V_2=\dfrac{1\times 20\times 546}{273\times 2}\\\\V_2=20~ml$

The volume does not change because the pressure and temperature are increased by the same ratio as the initial conditions (to 2x)

5 0

2 years ago

A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A

Pavel [41]

Answer:

92.49 %

Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and

From n = m/M

m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol

m = 0.0038 mol × 79.904 g/mol = 0.3036 g

% Br in compound = m₁/m₂ × 100%

m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)

m₂ = mass of compound = 0.3283 g

% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %

4 0

3 years ago