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2 years ago

Is marble a compound or element?

Chemistry

2 answers:

Vesna [10]2 years ago

7 0

Answer:

compound

Explanation:

"Marble is typically more than 95% calcium carbonate, perhaps even 99% calcium carbonate, and calcium carbonate is a compound. Marble has a good claim to be recognized as a compound."

vekshin12 years ago

6 0

Answer: Element

Explanation: No, marble is not a compound, because for a rock to be called marble other characteristics, beyond the chemical composition, have to have specific values. But marble is made up of only one compound, calcium carbonate. Colored marbles do contain small amounts of other chemicals.

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Which coefficient before potassium (K) balances this chemical equation?

miskamm [114]

Answer:

The coefficient before potassium (K) balances this chemical equation is 2.

Explanation:

_K +Cl₂ → 2KCl

K =1 ; Cl =2

K=1 × 2 = 2

Cl = 1 × 2 = 2

2 K +Cl₂ = 2 KCl

5 0

3 years ago

Read 2 more answers

CH4 + 202 → CO2 + 2H2O How many grams of O2 needed to produce 36 grams of H2O?

Kipish [7]

<h3>Answer:</h3>

64 g O₂

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

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 36 g H₂O

[Solve] x g O₂

Step 2: Identify Conversions

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Stoichiometry

Set up conversion: $\displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})$
Divide/Multiply [Cancel Units]: $\displaystyle 63.929 \ g \ O_2$

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

63.929 g O₂ ≈ 64 g O₂

8 0

2 years ago

Read 2 more answers

what is the balanced equation when copper metal is placed in a solution when platnium ii chloride is placed. what is the equatio

koban [17]

Answer:

$Cu~+~PtCl_2->Pt~+~CuCl_2$

Explanation:

In this case, we can start with the formula of Platinum (II) Chloride. The cation is the atom at the left of the name (in this case $Pt^+^2$ ) and the anion is the atom at the right of the name (in this case $Cl^-$ ). With this in mind, the formula would be $PtCl_2$ .