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pochemuha
3 years ago
14

Consider the total ionic equation below.

Chemistry
2 answers:
Tpy6a [65]3 years ago
4 0

Answer:

The net ionic equation is

Ba²⁺ + CO₃²⁻ → BaCO₃(s)

Explanation:

The chemical reaction described is given as

Ba²⁺ + 2NO₃⁻ + 2Na⁺ + CO₃²⁻ → BaCO₃(s) + 2Na⁺ + 2NO₃⁻

The net ionic equation is obtained by eliminating the ions that appear on both sides of the equation.

And those ions are 2Na⁺ and 2NO₃⁻.

This is a double displacement reaction in which BaNO₃ (aq) reacts with Na₂CO₃ (aq) to form BaCO₃ (s) and NaNO₃ (aq).

The aqueous solutions exist in ionic forms in the reaction and in the net ionic equation, ioms that appear on both sides of the overall equation obviously cancel out.

Hope this Helps!!!

Alexeev081 [22]3 years ago
4 0

Answer:

The correct answer is Ba2+ +2NO-3 =Ba(NO3)2

Explanation:

did it on edu

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How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?
Y_Kistochka [10]

There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

C_{6}H_{8}O_{6}

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g

To know the percent of each element, we need to to the following:

C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)

So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

4.58 grams of H

54.49 grams of O

We know the molecular masses of each element:

C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

C=\frac{40.92g}{12.010\frac{g}{mol}}=3.41mol\\\\H=\frac{4.58g}{1.008\frac{g}{mol}}=4.54mol\\\\O=\frac{54.49g}{15.999\frac{g}{mol}}=3.40mol

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

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How many moles are in 7.2x10^15 atoms of Pb?​
fiasKO [112]
<h3>Answer:</h3>

1.2 × 10⁻⁸ mol Pb

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
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<u>Chemistry</u>

<u>Atomic Structure</u>

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<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 7.2 × 10¹⁵ atoms Pb

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 7.2 \cdot 10^{15} \ atoms \ Pb(\frac{1 \ mol \ Pb}{6.022 \cdot 10^{23} \ atoms \ Pb})
  2. [DA] Multiply/Divide [Cancel out units]:                                                            \displaystyle 1.19562 \cdot 10^{-8} \ mol \ Pb

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.19562 × 10⁻⁸ mol Pb ≈ 1.2 × 10⁻⁸ mol Pb

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