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3 years ago

How many grams of sodium are needed to react completely with 19

Chemistry

1 answer:

gtnhenbr [62]3 years ago

6 0

Answer:

46 g of sodium

Explanation:

Sodium reacts vigorously with fluoride gas to form NaF as shown

2Na (s) + F2 (g) ------> 2NaF (s) (Na = 23, F = 19)

2 moles of Na reacts with 1 mole of F2 to produce NaF

By calculating the molar mass of the elements involved in the reaction then multiplied by the mole, the mass can be obtained.

23 * 2 g/mol of Na reacts with 1 * 19 g/mol of F2

46 g/mol of Na reacts with 19 g/mol of F2 to produce NaF

Since the mole ratio is 2 to 1 and 19 g of F2 is used for the reaction, 46 g of sodium will be consumed for the reaction to be achieved.

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How many atoms of iodine are in 12.75g of CaI2

dexar [7]

Answer:

0.5 × 10²³ atoms of iodine

Explanation:

Given data:

Mass of calcium iodide = 12.75 g

Number of atoms of iodine = ?

Solution:

First of all we will calculate the number of moles of calcium iodide.

Number of moles = mass/ molar mass

Number of moles = 12.75 g/ 293.9 g/mol

Number of moles = 0.04 mol

In one mole of calcium iodide there are two moles of iodine.

Thus in 0.04 moles:

0.04 mol × 2 = 0.08 moles of iodine

Now we will use the Avogadro number:

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.

1 mole = 6.022 × 10²³ atoms

0.08 moles of iodine × 6.022 × 10²³ atoms / 1 mol

0.5 × 10²³ atoms of iodine.

8 0

4 years ago

How do lipids differ from carbohydrates? they are formed from different basic elements

grin007 [14]

Lipids are insoluble in water and they produce twice as more energy as carbohydrates.

5 0

4 years ago

Write and balance the equation for the combustion of the fatty acid lauric acid, (C12H24O2).

Harlamova29_29 [7]

C12H24O2 +17 O2-------->12CO2 + 12H2O

<h3>Combustion:-</h3>

combustion is a chemical reaction that often involves the presence of oxygen and produces heat and light in the form of flames.

<h3>Lauric acid:-</h3>

Lauric acid has a 12-carbon backbone and is a saturated medium-chain fatty acid. In addition to being a key component of coconut oil and palm kernel oil, lauric acid occurs naturally in a variety of plant and animal fats and oils.

White solid lauric acid has a little bay oil odour to it.

Lauric acid is a cheap, non-toxic, and easy-to-handle substance that is frequently employed in lab studies on melting-point depression. Because lauric acid is a solid at ambient temperature but a liquid at boiling temperatures, it can be used to test different solutes to determine their molecular weights.

To learn more about Fatty acids refer to :-

brainly.com/question/26353151

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4 0

2 years ago

The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.

olchik [2.2K]

Answer:

Approximately $0.291\; \rm M$ (rounded to two significant figures.)

Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

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3 years ago

How do I balance chemical equations ?<br>

Maru [420]

Answer & explanation:

To balance a chemical equation, we must make sure that there are equal amounts of each element in the equation on either side.

To balance an equation step-by-step we first start by counting the amount of each element on each side. If they aren't equal, we must make them amounts equal on either side.

After you make sure all the elements are balanced, the net charge on both sides must also be equal.

8 0

4 years ago