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When a 5.00-g metal piece, A, was immersed in 38.0 mL of water, the water level rose to 50.0 mL. Similarly, when a 5.00-g metal

Mnenie [13.5K]

Answer:

A is denser than B as it's volume for the same mass is smaller.

Explanation:

Hello.

In this case, we first need to take into account that the density of each metal A and B is computed by dividing the mass over the volume of each metal which is actually computed by substracting the volume of water from the volume of the water and the solid:

$V_A=50.0mL-38.0mL=12.0mL\\\\V_B=60.0mL-38.0mL=22.0mL$

Next, we compute the densities as shown below:

$\rho_A=\frac{m_A}{V_A}=\frac{5.00g}{12.0mL}=0.42g/mL\\ \\\rho_B=\frac{m_B}{V_B}=\frac{5.00g}{22.0mL}=0.23g/mL$

In such a way, A is denser is B as it's volume for the same mass is smaller.

Best regards.

4 0

3 years ago

Aluminum reacts with chlorine gas to produce aluminum chloride according to the following equation. Al + Cl2 → AlCl3 Which of th

anzhelika [568]

<h3><u>Answer</u>;</h3>

B. 3/2

<h3><u> Explanation;</u></h3>

Balance the chemical equation

2Al + 3Cl2 → 2AlCl3

We want to convert moles of AlCl3 to moles of Cl2

The conversion factor is 2 mol AlCl3/3 mol Cl2.

We choose the one that makes the units cancel:

x mol AlCl3 x (3 mol Cl3)/(2mol AlCl3) = x mol Al

The fraction for the molar ratio is 3/2.

4 0

3 years ago

Read 2 more answers

In the reaction of silver nitrate with copper metal, metallic silver comes out of solution, and the solution turns blue. This as

nlexa [21]

In the reaction of silver nitrate with copper metal, metallic silver comes out of solution, and the solution turns blue. This as a <u>single replacement</u> reaction.

<h3>What is single replacement reaction?</h3>

A single replacement reaction, also known as a single displacement reaction, occurs when one element in a molecule is swapped out for another. The starting materials are always pure elements, such as a pure zinc metal or hydrogen gas, plus an aqueous compound.

A + BC → B + AC

When A is more reactive than B or when the product AC is more stable than BC, single replacement reactions happen. A and B could either be two halogens or two metals (with hydrogen included) (C is a cation). C functions as a spectator ion when BC and AC are in aqueous solutions.

For example, 2HCl(aq)+Zn(s)→ZnCl₂(aq)+H₂(g)

Learn more about single replacement reactions here:

brainly.com/question/19068047

#SPJ4

8 0

2 years ago

Identify the intermolecular attractions for dimethyl ether and for ethyl alcohol. Which molecule is expected to be more soluble

zheka24 [161]

Answer:

See explanation

Explanation:

All molecules possess the London dispersion forces. However London dispersion forces is the only kind of intermolecular interaction that exists in nonpolar substances.

So, the only kind of intermolecular interaction that exists in dimethyl ether is London dispersion forces.

As for ethyl alcohol, the molecule is polar due to the presence of polar O-H bond. In addition to London dispersion forces, dipole-dipole interactions and specifically hydrogen bonding also occurs between the molecules.

Because ethyl alcohol is polar, it is more soluble in water than dimethyl ether.

3 0

3 years ago

Which of these half-reactions represents reduction?

gogolik [260]

Answer: The half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element takes place is called reduction-half reaction.

For example, the oxidation state of Cr in $Cr_{2}O^{2-}_{7}$ is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

$Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}$

Similarly, oxidation state of Mn in $MnO^{-}_{4}$ is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

$MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}$

Thus, we can conclude that half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

3 0

3 years ago