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2 years ago

Cu2o+C➡Cu+Co2 what is the balance chemical equation

Chemistry

1 answer:

denis23 [38]2 years ago

5 0

<span>4 Cu2O + C2 = 8 Cu + 2 CO2 </span><span>Reaction type: single replacement i hope this hopes

</span>

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10. Chromium (III) oxide reacts with carbon dioxide to form..

Jobisdone [24]

Answer: chromium(III)

Explanation:

5 0

2 years ago

2. What is the volume of 40.0 g of N2 at STP?

vampirchik [111]

Answer:

Molar mass of

28 g/mol

Moles of

40.5 g N

gas

40.5 g

28 g/mol

1.45 mol

1 mol

of any gas at

STP

occupies

22.4 L

of volume.

1.45 mol

gas will occupy

1.45

mol

22.4 L

mol

32.48 L

3 0

2 years ago

Which is the correct number of moles of NO that is produced from 13.2 moles of oxygen

Oksanka [162]

<h3>Answer:</h3>

10.6 mol NO

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 4NH₃ + 5O₂ → 4NO + 6H₂O

[Given] 13.2 mol O₂

<u>Step 2: Identify Conversions</u>

[RxN] 5 mol O₂ → 4 mol NO

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 13.2 \ mol \ O_2(\frac{4 \ mol \ NO}{5 \ mol \ O_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 10.56 \ mol \ NO$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

10.56 mol NO ≈ 10.6 mol NO

5 0

2 years ago

Look at the position of Location 1, Location 2, and Location 3 in this picture. Simple diagram of a lake represented by a wavy c

alexgriva [62]

Answer: Location 3 is warmer than location 2 and 1

Explanation:

5 0

2 years ago

Read 2 more answers

Benzene is a minor component of gasoline. The standard molar enthalpy of formation of benzene C7H16(l) is 48.95 kJ/mol. For the

Radda [10]

Answer:

-3135.47 kJ/mol

Explanation:

Step 1: Write the balanced equation

C₆H₆(l) + 7.5 O₂(g) ⇒ 6 CO₂(g) + 3 H₂O(g)

Step 2: Calculate the standard enthalpy change of the reaction (ΔH°r)

We will use the following expression.

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpies of formation

p: products

r: reactants

ΔH°r = 6 mol × ΔH°f(CO₂(g)) + 3 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₆H₆(l)) - 7.5 mol × ΔH°f(O₂(g))

ΔH°r = 6 mol × (-393.51 kJ/mol) + 3 mol × (-241.82 kJ/mol) - 1 mol × (48.95 kJ/mol) - 7.5 mol × 0 kJ/mol

ΔH°r = -3135.47 kJ

Since this enthalpy change is for 1 mole of C₆H₆(l), we can express it as -3135.47 kJ/mol.

7 0

2 years ago