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son4ous [18]
2 years ago
14

Competition occurs when

Chemistry
1 answer:
Katen [24]2 years ago
3 0

Answer:

Competition occurs when <em>(A) two or more organisms need the same resource.</em>

Explanation:

I dont think it could be (B), (C) or (D) because those don't really make sense.

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The total number of sodium atoms in 46.0 grams of sodium<br> is
marusya05 [52]
The answer is 2.0, and it is because
3 0
3 years ago
What volume of 10,0 M H2SO4 is required to prepare 4.0 L of 0.50 M H2SO4?
Evgesh-ka [11]

Answer: A)  0.20 L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

According to the neutralization law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock H_2SO_4 solution = 10.0 M

V_1 = volume of stock H_2SO_4   solution = ?

M_2 = molarity of dilute H_2SO_4  solution = 0.50 M

V_2 = volume of dilute H_2SO_4  solution = 4.0 L

10.0\times V_1=0.50\times 4.0

V_1=0.20L

Therefore, the volume of 10.0 M H_2SO_4 solution needed to prepare 4.0 L of 0.50 M H_2SO_4 is 0.20 L

8 0
3 years ago
What are examples of phenotypes
Naddika [18.5K]

Answer:

Explanation:

Examples of phenotypes include height, wing length, and hair color. Phenotypes also include observable characteristics that can be measured in the laboratory, such as levels of hormones or blood cells.

3 0
3 years ago
Read 2 more answers
What is the product of the reaction of (S)-2-bromobutane with sodium methoxide in acetone?
Serggg [28]

Answer:

2-methoxybutane

Explanation:

This reaction is an example of Nucleophilic substitution reaction. Also, the reaction of (S)-2-bromobutane with sodium methoxide in acetone, is bimolecular nucleophilic substitution (SN2). The reaction equation is given below.

(S)-2-bromobutane + sodium methoxide (in acetone) → 2-methoxybutane

7 0
3 years ago
aseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.5 g of methane is
maksim [4K]

Answer:

There is 9.6 grams of CO2 produced

Explanation:

Step 1: Data given

Mass of methane = 5.50 grams

Molar mass of methane = 16.04 g/mol

Mass of oxygen = 13.9 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The reaction

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles methane = 5.50 grams / 16.04 g/mol

Moles methane = 0.343 moles

Moles oxygen = 13.9 grams / 32.0 g/mol

Moles oxygen = 0.434 moles

For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

O2 is the limiting reactant. It will completely react (0.434 moles).

There will react 0.434/2 = 0.217 moles CH4

There will remain 0.343-0.217 = 0.126 moles CH4

There will be produced 0.434 moles of H2O and

0.434/2 =0.217 moles of CO2

Step 4: Calculate mass of products

Mass = moles * molar mass

Mass CO2 = 0.217 moles ¨44.01 g/mol

Mass CO2 = 9.6 grams

Mass H2O = 0.434 moles * 18.02

Mass H2O = 7.8 grams

4 0
3 years ago
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