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Dominik [7]
3 years ago
6

The cubic centimeter (cm3) is a measurement of which of the following quantities?

Chemistry
2 answers:
xz_007 [3.2K]3 years ago
6 0
Volume I’m pretty sure
marishachu [46]3 years ago
5 0
Volume is measured by cm3 bc itconsist of length ×width ×height
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How many moles of chromium III nitrate are produced When chromium reacts with 0.85 moles of lead for nitrate to produce chromium
Pepsi [2]

0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.

<h3>Explanation</h3>

Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as \text{Pb}^{2+} ions. The formula for a nitrate ion is {\text{NO}_3}^{-}. The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each \text{Pb}^{2+} ion in lead nitrate would pair up with two {\text{NO}_3}^{-} ions. The formula for lead nitrate will be \text{Pb}({\text{NO}_3})_2. Each formula unit of lead nitrate will contain one \text{Pb}^{2+} ion and two {\text{NO}_3}^{-} ions.

The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as \text{Cr}^{3+}. Similarly, each \text{Cr}^{3+} will pair up with three {\text{NO}_3}^{-} ions. The formula for chromium (III) nitrate will be \text{Cr}(\text{NO}_3})_3. Each formula unit of chromium (III) nitrate will contain one {\text{NO}_3}^{-} ion and three {\text{NO}_3}^{-} ions.

0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of {\text{NO}_3}^{-} ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.

7 0
3 years ago
O tetraclorometano é um solvente muito utilizado na indústria de transformaç?o de produtos orgânicos ele é produzido pela reaç?o
Vlada [557]

Answer:

ΔHr = -103,4 kcal/mol

Explanation:

<u>Using:</u>

<u>AH° (kcal/mol) </u>

<u>Metano (CH) </u>

<u>-17,9 </u>

<u>Cloro (CI) </u>

<u>tetraclorometano (CCI) </u>

<u>- 33,3 </u>

<u>Acido cloridrico (HCI) </u>

<u>-22</u>

It is possible to obtain the ΔH of a reaction from ΔH's of formation for each compound, thus:

ΔHr = (ΔH products - ΔH reactants)

For the reaction:

CH₄(g) + Cl₂(g) → CCl₄(g) + HCl(g)

The balanced reaction is:

CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g)

The ΔH's of formation for these compounds are:

ΔH CH₄(g): -17,9 kcal/mol

ΔH Cl₂(g): 0 kcal/mol

ΔH CCl₄(g): -33,3 kcal/mol

ΔH HCl(g): -22 kcal/mol

The ΔHr is:

-33,3 kcal/mol × 1 mol + -22 kcal/mol× 4 mol - (-17,9 kcal/mol × 1 mol + 0kcal/mol × 4mol)

<em>ΔHr = -103,4 kcal/mol</em>

<em></em>

I hope it helps!

3 0
3 years ago
Calculate the mass of 488 moles of calcium carbonate
san4es73 [151]

Answer: 48800g

Explanation:

Using the mathematical relation : Moles = Mass / Molar Mass

Moles = 488

Molar mass of CaCO3 = 40 + 12 + (16 x 3) = 100g/mol

Therefore

488 = mass / 100 = 48800g

6 0
3 years ago
If something has an actual yield of 2.5 grams and a theoretical yield of 7.5 grams, what is the percent yield?
aivan3 [116]
2.5/7.5 = 0. 3333 * 100% = 33.33%
8 0
3 years ago
A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )
Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g)   (1)

The reaction rate of equation (1) is given by:

rate = k*[O_{3}][NO]     (2)

<u>We have:</u>

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s  

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

\frac{\Delta[NO_{2}]}{\Delta t} = rate

\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!                                

5 0
3 years ago
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