Explain why Alloys are harder than pure metals.
b. Drill heads also contain diamonds.
Describe as fully as you can the structure and bonding in diamond.
c. Polymers are produced from crude oil.
Describe the structure and bonding in thermosoftening polymer and explain why thermosoftening polymers melt when heated.
In this reaction, ammonia molecules (NH3) act as a base because they
(1) accept hydrogen ions (H+)
(2) accept hydroxide ions (OH-)
(3) donate hydrogen ions (H+)
(4) donate hydroxide ions (OH-)
Answer:
1. accept hydrogen ions (H+)
Explanation:
In the given reaction, NH3 acts as a base because it accepts hydrogen ions (H+) from the acid HCl to form the conjugate acid NH4+. Therefore, the correct answer is 1. accept hydrogen ions (H+).
Answer: Option (4) is the correct answer.
Explanation:
Activation energy is the minimum amount of energy required by reactant molecules to undergo a chemical reaction.
Whereas a catalyst is defined as the substance that helps in increasing the rate of reaction by decreasing the activation energy without itself getting consumed in the reaction.
When a catalyst decreases the activation energy then molecules with lesser energy become able to participate in the reaction and thus, products are obtained at a faster rate. Hence, a catalyst increases the rate of a reaction.
Thus, we can conclude that a catalyst works by decreasing the activation energy required for a reaction.
Answer:
11.0 dm³
Explanation:
From the question,
Applying
PV= nRT............... Equation 1
Where P = pressure of oxygen gas, V = volume of oxygen gas, n = number of moles of oxygen, R = molar constant, T = Temperature.
make V the subeject of the equation
V = nRT/P............. Equation 2
But,
Number of mole (n) = Mass of oxygen(m)/Molar mass of oxygen(m')
n = m/m'....................... Equation 3
Substitute equation 3 into equation 2
V = mRT/Pm'............. Equation 4
Given: T = 28°C = (28+273) = 301 K, P = 0.998 torr = (0.998×0.00131579) = 1.3132 atm, m = 18.4 g
Constant: R = 0.082 atm.dm³/K.mol, m' = 32 g/mol.
Substitute these values into equation 4
V = (301×18.4×0.082)/(32×1.3132)
V = 454.1488/42.0224
V = 10.81 dm³
V = 11.0 dm³
parasitism
B.
amensalism
C.
commensalism
D.
symbiosis
Answer: C: commensalism please give me hearts and stars because the answer is right!!
Explanation:
The relationship between the clown fish and the deadly sea anemone can be classified as commensalism. Though the sea anemone’s tentacles are covered with thousands of nematocysts (tiny, venomous harpoons), the clown fish is protected by a thick coat of mucus and is not harmed by the anemone’s venom. The clown fish is therefore protected from predators and is also able to get leftover food from the anemone's meals. Since the clown fish ultimately benefits from this arrangement and the sea anemone neither benefits nor suffers because of the clown fish, the relationship is commensalistic.
The clown fish and the sea anemone exhibit a mutualistic relationship, where both organisms benefit. However, among the options provided in the question, the most accurate is 'symbiosis', which broadly denotes a close, long-term interaction between different biological species.
The symbiotic relationship between the clown fish and the sea anemone is referred to as mutualism. This term, which is missing from the options, indicates a relationship in which both organisms benefit. The clown fish is protected from predators and has access to food remnants, whereas the sea anemone gets cleaned by the clown fish, which removes parasites and debris. Among the options provided, the closest would be symbiosis (option D), as it describes a close, long-term interaction between different biological species, which can include mutualism, commensalism, and parasitism.
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To produce 100.0 L of NH3 at STP, 62.4 grams of N2 are required.
The balanced equation for the reaction is:
N2(g) + 3H2(g) → 2NH3(g)
According to the balanced equation, one mole of N2 reacts with three moles of H2 to produce two moles of NH3. From this information, we can use stoichiometry to determine the mass of N2 required to produce 100.0 L of NH3 at STP.
First, we need to convert liters of NH3 to moles using the ideal gas law and the molar volume of a gas at STP (~22.4 L/mol). Once we have the moles of NH3, we can use the mole ratio from the balanced equation to calculate the moles of N2. Finally, we can use the molar mass of N2 to convert moles to grams.
Let's calculate:
Therefore, 62.4 grams of N2 are required to produce 100.0 liters of NH3 at STP.
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