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2 years ago

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A cube of an unknown metal measures 0.200 cm on one side. The mass of the cube is 52 mg. Which of the following is most likely t

he unknown metal?
Metal Density(g/cm^3)
rhodium 12.4
copper 8.96
niobium 8.57
vanadium 6.11
zirconium 6.51

a. copper
b. rhodium
c. niobium
d. vanadium
e. zirconium

1 answer:

Margarita [4]2 years ago

7 0

Answer:

Option E. Zirconium

Explanation:

From the question given above, the following data were obtained:

Length of side (L) of cube = 0.2 cm

Mass (m) of cube = 52 mg

Name of the unknown metal =?

Next, we shall determine the volume of the cube. This can be obtained as follow:

Length of side (L) of cube = 0.2 cm

Volume (V) of the cube =?

V = L³

V = 0.2³

V = 0.008 cm³

Next, we shall convert 52 mg to g. This can be obtained as follow:

1000 mg = 1 g

Therefore,

52 mg = 52 mg × 1 g / 1000 mg

52 mg = 0.052 g

Thus, 52 mg is equivalent to 0.052 g.

Next, we shall determine the density of the unknown metal. This can be obtained as follow:

Mass = 0.052 g.

Volume = 0.008 cm³

Density =?

Density = mass / volume

Density = 0.052 / 0.008

Density of the unknown metal = 6.5 g/cm³

Comparing the density of the unknown metal i.e 6.5 g/cm³ with those given in table in the above, we can conclude that the unknown metal is zirconium

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Answer:

The molar mass of the unknown acid is 89 g/mol

Explanation:

<u>Step 1:</u> The balanced equation

HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)

It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.

<u>Step 2:</u> Data given

Mass of the acid = 0.093 grams

volume = 250 mL

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adds 6.5 mL NaOH

<u>Step 3: </u>Calculate moles of NaOH

We know the concentration and volume of NaOH needed to neutralize the acid.

By determining the moles of NaOH in that volume in liters (95.9mL=0.0959L), the moles of acid in the original sample can be determined from the reaction stoichiometry.

Moles = Molarity * Volume

Moles = 0.16 M * 0.0065 L

Moles = 0.00104 moles NaOH

<u>Step 4: </u>Calculate moles of the unknown acid:

It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.

For 0.00104 moles NaOH we have 0.00104 moles of HA

<u>Step 5: </u>Calculate the molar mass of the acid

Molar mass Ha = Mass Ha / moles Ha

Molar mass Ha = 0.093 grams / 0.00104 moles

Molar mass Ha = 89.42 g/mol ≈89 g/mol

The molar mass of the unknown acid is 89 g/mol

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Answer:

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Explanation:

Bond length is inversely related to bond strength. The longer the bond length, the weaker the bond. The shorter the bond length the stronger the bond. A large bond distance implies that there is poor interaction between the atoms involved in the bond. A long bond distance or bond length may even indicate the absence of covalent interaction between the atoms involved.

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An oily liquid with a cheesy, waxy odor like that of goats or other barnyard animals is caproic acid, a compound containing carb

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The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O

We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:

How to determine the mass of C

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Mass of C =?

Mass of C = (12 / 44) × 9.78

Mass of C = 2.67 g

How to determine the mass of H

Mass of H₂O = 20.99 g
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Mass of H =?

Mass of H = (2 / 18) × 4

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How to determine the mass of O

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Mass of O = (mass of compound) – (mass of C + mass of H)

Mass of O = 4.30 – (2.67 + 0.44)

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<h3>How to determine the empirical formula </h3>

The empirical formula of the compound can be obtained as follow:

C = 2.67 g
H = 0.44 g
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Divide by their molar mass

C = 2.67 / 12 = 0.2225

H = 0.44 / 1 = 0.44

O = 1.19 / 16 = 0.074

Divide by the smallest

C = 0.2225 / 0.074 = 3

H = 0.44 / 0.0744 = 6

O = 0.074 / 0.074 = 1

Thus, the empirical formula of the compound is C₃H₆O

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1 year ago

In the gas-phase reaction 2 A + B ⇌ 3 C + 2 D, it was found that, when 1.00 mol A, 2.00 mol B, and 1.00 mol D were mixed and all

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Answer:

(i)The mole fractions are :

$A=\frac{0.4}{4.3} \\=0.0930$

$B=\frac{1.4}{4.3} \\=0.3256$

$C=\frac{0.9}{4.3} \\=0.2093$

$D=\frac{1.6}{4.3} \\=0.3721$

(ii) $K=0.4508$

(iii)ΔG = 1.974kJ

Explanation:

The given equation is :

$2A+B$ ⇄ $3C+2D$

Let $\alpha$ be the number of moles dissociated per mole of $B$

Thus ,

<em>The initial number of moles of :</em>

$A=1$

$B=2$

$C=0$

$D=1$

$2A\\(1-2\alpha)$ + $B\\2(1-\alpha)$ ⇄ $3C\\(3\alpha)$ + $2D\\(1+2\alpha)$

And finally the number of moles of $C[tex] is 0.9Thus ,[tex]3\alpha=0.9\\\alpha=0.3[tex]The final number of moles of:[tex]A = 1-2\alpha=1-2*0.3=0.4mol[tex] [tex]B=2(1-\alpha)=2(1-0.3)=1.4mol[tex][tex]D=1+2\alpha=1+2*0.3=1.6mol[tex]Thus , total number of moles are : 0.4+1.4+0.9+1.6=4.3(i)The mole fractions are : [tex]A=\frac{0.4}{4.3} \\=0.0930$

$B=\frac{1.4}{4.3} \\=0.3256$

$C=\frac{0.9}{4.3} \\=0.2093$

$D=\frac{1.6}{4.3} \\=0.3721$

(ii)

$K=\frac{(P_C^3)(P_D^2)}{(P_A^2)(P_B)}$

Where ,

$P_A,P_B,P_C,P_D$ are the partial pressures of A,B,C,D respectively.

Total pressure = 1 bar .

∴

<em> $P_A= 0.0930*1=0.0930$ </em>

<em> $P_B= 0.3256*1=0.3256$ </em>

<em> $P_C= 0.2093*1=0.2093$ </em>

<em> $P_D= 0.3721*1=0.3721$ </em>

$K=\frac{0.2093^3*0.3721^2}{0.0930^2*0.3256} \\K=0.4508$

(iii)

Δ $G=-RTlnK\\$

ΔG = $-8.314*(273+25)*ln(0.4508)\\=1973.96J\\=1.974kJ$

3 0

2 years ago