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trasher [3.6K]
3 years ago
6

What is one thing that is the same about a mole of sodiums and a mole of carbons?

Chemistry
1 answer:
sweet-ann [11.9K]3 years ago
7 0
C the total number of atoms
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What are the products of the balanced equation for the combustion of C8H17OH ?
erastova [34]
C8H17OH + 12O2 —> 8CO2 + 9H2O
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3 years ago
Convert 7.1X1025 molecules of water to moles
scoundrel [369]

Answer:

Moles of water in 7.1×10²⁵ molecules are 118 mol.

Explanation:

Given data:

Number of molecules of water = 7.1×10²⁵ molecules

Moles of water in 7.1×10²⁵ molecules = ?

Solution;

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

1 mole = 6.022 × 10²³ molecules of water

7.1×10²⁵ molecules of water × 1 mol / 6.022 × 10²³ molecules of water

1.18×10² moles of water 0r 118 moles of water

5 0
3 years ago
Using off-road vehicles does not contribute to the process of erosion. true or false
Artemon [7]
Using off road vehicles does help contribute to the process of erosion.
7 0
3 years ago
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150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac
Murljashka [212]

Answer:

0.175\; \rm mol \cdot L^{-1}.

Explanation:

Magnesium chloride and silver nitrate reacts at a 2:1 ratio:

\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s).

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to  \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}.

The precipitate silver chloride \rm AgCl is insoluble in water and barely ionizes. Hence, \rm AgCl\! isn't rewritten as ions.

Net ionic equation:

\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}.

Calculate the initial quantity of nitrate ions in the mixture.

\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}.

Since nitrate ions \rm {NO_3}^{-} do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol.

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be (1/2) of the concentration in the original solution.

\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}.

6 0
3 years ago
Which of the following measurements has three significant figures?
Free_Kalibri [48]

The correct answer is <u>C: 0.510 m</u>

<u></u>

N I C E - D A Y!

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2 years ago
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