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3 years ago

Calculate the density of a block of metal with a volume of 12.5 cm3 and mass of 126.0 g

Chemistry

1 answer:

galina1969 [7]3 years ago

3 0

Answer:

This is math. by the way read further

So density is equal mass/volume

The units are OK we don't need to change them.

D=126/12.5=10.08 grams per cm^3

Answer 10.08 g/cm^3

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How many moles do you have in 37.3 g of Co(CrO4)3

shutvik [7]

Answer:

0.0917 mol Co(CrO₄)₃

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis

Explanation:

Step 1: Define

37.3 g Co(CrO₄)₃

Step 2: Identify Conversions

Molar Mass of Co - 58.93 g/mol

Molar Mass of Cr - 52.00 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Co(CrO₄)₃ - 58.93 + 3(52.00) + 12(16.00) = 406.93 g/mol

Step 3: Convert

$37.3 \ g \ Co(CrO_4)_3(\frac{1 \ mol \ Co(CrO_4)_3}{406.93 \ g \ Co(CrO_4)_3} )$ = 0.091662 mol Co(CrO₄)₃

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

0.091662 mol Co(CrO₄)₃ ≈ 0.0917 mol Co(CrO₄)₃

8 0

2 years ago

If 4.80 mol Ca mixed with 2 mol N2, which is the limiting reactant? 3Ca (s) + N2 (g) Ca3N2 (s)

Andreyy89

Nitrogen in the limiting reactant x

4 0

2 years ago

The C=C bond in 2-cyclohexenone produces an unusually strong signal. Explain using resonance structuresBased on the above resona

frosja888 [35]

Answer:

When two single single bonds separated by a double bond (e.g C=C-C=C or C=C-C=O in the case of 2-cyclohexenone), the effect of resonance among those there bonds will be observed.

Explanation:

Since the Oxygen atom has higher electronegativity, it will cause the electrons in the resonance bonds 'flow' toward the Oxygen atom, so that the C=C will 'lose' some electron. The signal read for that bond will be different from other alkene structure.

Attachment is the resonance structure of 2-cyclohexene.

8 0

3 years ago

Chemistry table practice

Sonbull [250]

Explanation:

I don't understand your question....

thank you

5 0

3 years ago

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.