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3 years ago

Which statement best compares segmented worms and roundworms?

Chemistry

2 answers:

Mariana [72]3 years ago

4 0

Answer:

I think it's A.

Explanation:

Metameric bodies are divided into segments called annuli. Annuli are covered in tiny hairs called setae. Setae help the worms move. So yes it's A.

<em>Happy Thanksgiving too! :D</em>

Rudik [331]3 years ago

4 0

Pretty sure it’s A- i think

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A compound is found to contain 50. 05% sulfur and 49. 95% oxygen by mass. What is the empirical formula for this compound? SO S2

jekas [21]

The empirical formula for the given compound has been $\rm SO_2$ . Thus, option C is correct.

The empirical formula has been the whole unit ratio of the elements in the formula unit.

<h3>Computation for the Empirical formula</h3>

The given mass of Sulfur has been, 50.05 g

The given mass of oxygen has been 49.95 g.

The moles of elements in the sample has been given by:

$\rm Moles=\dfrac{Mass}{Molar\;mass}$

Moles of Sulfur:

$\rm Moles\;S=\dfrac{50.05}{32}\\
 Moles\;S=1.56\;mol$

The moles of sulfur in the unit has been 1.56 mol.

Moles of Oxygen:

$\rm Moles\;O=\dfrac{49.95}{16} \\
Moles\;O=3.12\;mol$

The moles of oxygen in the unit has been 3.12 mol.

The empirical formula unit has been given as:

$\rm S_{1.56}O_{3.12}=SO_2$

Thus, the empirical formula for the given compound has been $\rm SO_2$ . Thus, option C is correct.

Learn more about empirical formula, here:

brainly.com/question/11588623

8 0

2 years ago

Characteristics of a substance that can be measured or observed by changing the identity of the material is a

algol [13]

C cause ur changing the identity of the material

4 0

2 years ago

For the reaction between aqueous silver nitrate and aqueous sodium chloride, write each of the following. The products of the re

vova2212 [387]

Answer:

A balanced ionic equation shows the reacting ions in a chemical reaction. These equations can be used to represent what happens in precipitation reactions or displacement reactions.

Precipitation reactions

In a typical precipitation reaction, two soluble reactants form an insoluble product and a soluble product.

For example, silver nitrate solution reacts with sodium chloride solution. Insoluble solid silver chloride and sodium nitrate solution form:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

The Na+ ions and NO3- ions remain separate in the sodium nitrate solution and do not form a precipitate. Ions that remain essentially unchanged during a reaction are called spectator ions.This means these can be ignored when writing the ionic equation. Only how the solid silver chloride forms is needed to be shown:

Ag+(aq) + Cl-(aq) → AgCl(s)

In a balanced ionic equation:

the number of positive and negative charges is the same

the numbers of atoms of each element on the left and right are the same

Displacement reactions

Displacement reactions take place when a reactive element displaces a less reactive element from one of its compounds.

A common type of displacement reaction takes place when a reactive metal reacts with the salt of a less reactive metal. For example, copper reacts with silver nitrate solution to produce silver and copper(II) nitrate solution:

2AgNO3(aq) + Cu(s) → 2Ag(s) + Cu(NO3)2(aq)

In this reaction, the NO3- ions remain in the solution and do not react - they are the spectator ions in this reaction. So, they can be removed from the ionic equation:

2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)

Question

Explain why this ionic equation is balanced:

Ba2+(aq) + SO42-(aq) → BaSO4(s)

Hide answer

There are the same numbers of atoms of each element on both sides of the equation. The total charge on both sides is also the same (zero).

Question

Balance this ionic equation, which represents the formation of a silver carbonate precipitate:

Ag+(aq) + CO32-(aq) → Ag2CO3(s)

Hide answer

2Ag+(aq) + CO32-(aq) → Ag2CO3(s)

Question

Balance this ionic equation, which represents the displacement of iodine from iodide ions by chlorine:

Cl2(aq) + I-(aq) → I2(aq) + Cl-(aq)

Hide answer

Cl2(aq) + 2I-(aq) → I2(aq) + 2Cl-(aq

Explanation:

this will help, I used this for my work x

4 0

3 years ago

60 grams of ice will require _____ calories to raise the temperature 1°C.

guajiro [1.7K]

Answer:

60 grams of ice will require 30.26 calories to raise the temperature 1°C.

Explanation:

The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:

where, Q is the amount of heat released or absorbed by the system.

m is the mass of the ice (m = 60.0 g).

c is the specific heat capacity of ice (c = 2.108 J/g.°C).

ΔT is the temperature difference (ΔT = 1.0 °C).

∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.

<em>It is known that 1.0 cal = 4.18 J.</em>

7 0

3 years ago

Suppose the half-life of element k is 20 minutes. if you have a 200 g sample of k, how much of element k will be left after 60 m

vodka [1.7K]

T=20 min
m₀=200 g
t=60 min

the mass of element through time t is:
m=m₀*2^(-t/T)

m=200*2^(-60/20)=25 g

25 grams of element will be left after 60 minutes

5 0

3 years ago

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