Three different species of microscopic, thread-like worms are responsible for the parasitic disease lymphatic filariasis. Adult worms can only survive in the lymphatic system of humans.
What regions are affected by lymphatic filariasis?Nowadays, it is indigenous to Madagascar, many Western Pacific Island countries and territories, Sub-Saharan Africa (with the exception of the southernmost region of the continent), and parts of the Caribbean. South America, India, and Southeast Asia are also infrequent regions where Bancroftian filariasis can be found. Brugman spp.
What kind of mosquito carries dengue?by way of mosquito bites. The mosquito aedes aegypti. When an infected Aedes species mosquito (Ae. aegypti or Ae. albopictus) bites a human, the virus that causes dengue is transmitted.
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A gas at 300 k and 4. 0 atm is moved to a new location with a temperature of 250 k. The volume changes from 5. 5 l to 2. 0 l. What is the pressure of the gas at the new location? use the formula: p1v1 t1 = p2v2 t2
The pressure of the gas at the new location is approximately 7.92 atm.
To solve this problem using the formula provided, we need to plug in the given values for the initial state (denoted by subscript 1) and the final state (denoted by subscript 2) of the gas:
p₁v₁t₁ = p₂v₂t₂
where:
p₁ = 4.0 atm (initial pressure)
v₁ = 5.5 L (initial volume)
t₁ = 300 K (initial temperature)
p₂ = ? (what we are solving for)
v₂ = 2.0 L (final volume)
t₂ = 250 K (final temperature)
We can rearrange the equation to solve for p₂:
p₂ = p₁v₁t₁ / v₂t₂
Substituting the given values:
p₂ = (4.0 atm) x (5.5 L) x (300 K) / (2.0 L) x (250 K)
p₂ = 7.92 atm
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Answer:
its 9.2 atm
Explanation:
Ag2S forms a black precipitate and CdS forms a dark yellow precipitate. A solution of CdS is prepared by dissolving 0. 25 moles of solid CdS in distilled water. When equilibrium is established, a small amount of dark yellow solid remains on the bottom. A 0. 25 mol if solid AgNO3 is then stirred into the beaker, and the dark yellow precipitate is replaced with a black precipitate.
I. Did the following equilibrium shift after the AgNO3 was added? Explain using 2-3 sentences.
CdS (s) equilibrium arrows Cd 2+ (aq) + S 2- (aq)
II. Which compound has the larger Ksp value? Justify your answer
Yes, the equilibrium shifted after the AgNO₃ was added and Ag₂S having a larger Ksp value as compared to CdS.
The addition of AgNO₃ provides Ag⁺ ions, which react with S₂⁻ ions in the solution to form insoluble Ag₂S, causing the equilibrium to shift towards the products. This results in the formation of a black precipitate, indicating the presence of Ag₂S.
Ag₂S has a larger Ksp value compared to CdS. This is because Ag₂S has a lower solubility product constant (Ksp) than CdS, which means it is less soluble in water. The Ksp value for Ag₂S is 8.5 x 10⁻⁵¹, while the Ksp value for CdS is 4.0 x 10⁻²⁷. Therefore, Ag₂S has a larger Ksp value and is less soluble in water than CdS.
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If 1.90 g of silver are produced from the reaction, how many moles of copper(II) nitrate are also produced?
If 1.90 g of silver are produced from the reaction, then 0.0176 moles of copper(II) nitrate are also produced.
Why is a chemical reaction?A chemical reaction occurs when moving molecules hit each other, breaking their bonds and producing an exchange of atoms that form new products. Another way a chemical reaction can occur is through the vibration of substances; when they do so with sufficient energy, they can be broken down into smaller molecules.
we need to use the balanced chemical equation,
3Cu(NO3)2 + 2Al → 3Cu + 2Al(NO3)3
we can see that 3 moles of Cu(NO3)2 react with 2 moles of Al to produce 3 moles of Cu. Therefore, the molar ratio of Cu(NO3)2 to Cu is 3:3 or 1:1.
m(Ag) = 1.90 g
M(Ag) = 107.87 g/mol
n(Ag) = m(Ag) / M(Ag)
n(Ag) = 1.90 g / 107.87 g/mol
n(Ag) = 0.0176 mol
The molar ratio of Cu(NO3)2 to Cu is 1:1, the moles of Cu produced in the reaction is also 0.0176 mol. So, the number of moles of Cu(NO3)2 produced is also 0.0176 mol.
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which of the following solutions is a buffer? check all that apply. which of the following solutions is a buffer?check all that apply. a solution made by mixing 100 ml of 0.100 m hcook and 50 ml of 0.100 m kcl a solution made by mixing 100 ml of 0.100 m hcooh and 50 ml of 0.100 m hcl a solution made by mixing 100 ml of 0.100 m hcooh and 50 ml of 0.100 m naoh a solution made by mixing 100 ml of 0.100 m hcooh and 500 ml of 0.100 m naoh
The following solutions are buffers:
A solution made by mixing 100 ml of 0.100 M HCOOH and 50 ml of 0.100 M KCl, and a solution made by mixing 100 ml of 0.100 M HCOOH and 500 ml of 0.100 M NaOH.
Let's learn more about buffer solutions:
1. Buffer solutions can withstand or resist pH changes, even when adding a strong acid or base. A buffer solution comprises a weak acid and its conjugate base or a weak base and its conjugate acid. The pH change is insignificant when an acid or a base is added to a buffer solution.
2. The chemical equation for the ionization of HCOOH in water is as follows:
HCOOH + H2O ⇌ H3O+ + HCOO-
The salt made from the acid and the base in the buffer solution must be able to act as a source of ions to neutralize either the excess H3O+ ions or OH- ions introduced into the solution. For example, KCl and NaOH are used in the buffer solutions listed above, as they produce K+ and Na+ ions, which can interact with HCOO- ions in the first buffer and HCOO- ions and HCOOH molecules in the second buffer.
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