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borishaifa [10]

3 years ago

5

Write the cations and anions present in NaCl

1 answer:

pochemuha3 years ago

6 0

In table salt, sodium chloride, sodium is the cation (Na+) and chloride is the anion (Cl-).

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What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 5

charle [14.2K]

Answer:

The volume we need is 8.89 mL

Explanation:

We analyse data:

36 % by mass → 36 g of HCl in 100 g of solution

Solution's density = 1.179 g/mL

5.20L → Volume of diluted density

pH = 1.70 → [H⁺] = 10⁻¹'⁷⁰ = 0.0199 M

HCl → H⁺ + Cl⁻

0.0199

pH can gives the information of protons concentrations, so as ratio is 1:1, 0.0199 M is also the molar concentration of HCl

Let's verify the molar concentration of the concentrated solution:

We convert the mass to moles: 36 g / 36.45 g/mol = 0.987 moles

As the solution mass is 100 g, we apply density to find out the volume:

Density = Mass / volume → Volume = Mass / Density

Volume = 100 g / 1.179 g/mL → 84.8 mL

Let's convert the volume from mL to L in order to define molarity

84.8 mL . 1L/ 1000mL = 0.0848 L

Molarity → 0.987 mol / 0.0848L = 11.6M

Let's apply the dilution formula:

M concentrated . V concentrated = M diluted . V diluted

11.6 M . V concentrated = 0.0199M . 5.20L

V concentrated = (0.0199M . 5.20L) / 11.6M → 8.89×10⁻³L

We can say, that the volume we need is 8.89 mL

8 0

3 years ago

You add 10.0 grams of solid copper(II) phosphate, Cu3(PO4)2, to a beaker and then add 100.0 mL of water to the beaker at T = 298

Natalija [7]

Answer:

1.4×10-37

Explanation:

The equation for the dissolution of the copper II phosphate is first written as shown and the ICE table is set up as also shown. S is obtained as shown and this is now used to obtain 2s and subsequently the solubility product of the calcium phosphate as shown in detail in the image attached to this solution. The step-by-step solution shows how to obtain Ksp when concentration at equilibrium is given.

3 0

4 years ago

An anode is an electrode immersed in the half-cell where _____ takes place.

MrMuchimi

If I'm not mistaken, the answer should be B. Oxidation

5 0

4 years ago

Read 2 more answers

Chemical Quantities "Molar Mass - 2 step" - Wksh #3 Directions: Use dimensional analysis to perform the following calculations.

fomenos

There are about 1.27 × 10^23 molecules present in 13.5 g of SO2.

<h3>What are molecules?</h3>

A molecule is the simplest part of a compound that have independent existence.

Number of moles of sulfur dioxide = 13.5 g/64 g/mol = 0.211 moles

If 1 mole of SO2 contains 6.02 × 10^23 molecules

0.211 moles of SO2 contains 0.211 moles × 6.02 × 10^23 molecules/1 mole

= 1.27 × 10^23 molecules

32 g of sulfur contains 6.02 × 10^23 atoms

x g of sulfur contains 2.23 x 10^23

x = 32 × 2.23 x 10^23 /6.02 × 10^23

x = 11.85 g

6.02 × 10^23 formula units has a mass of AgF = 127 g

x formula units has a mass of = 42.15 g

x = 6.02 × 10^23 × 42.15/127

x = 2 × 10^23 formula units

6.02 × 10^23 formula units of Fe2O3 has a mass of 160 g

8.83 x 10^23 formula units of Fe2O3 has a mass of 8.83 x 10^23 x 160 g/6.02 × 10^23

x = 235 g

Learn more about formula unit: brainly.com/question/19293051

6 0

2 years ago

What mass of carbon dioxide is produced from the complete combustion of 6.00×10−3 g of methane?

Crazy boy [7]

The answer for the following problem is explained below.

<u><em>Therefore 16.5 × 10^-3 grams of carbon dioxide is produced from the complete combustion.</em></u>

Explanation:

Given:

mass of methane = 6.00 × 10^-3 grams

$CH_{4}$ + $O_{2}$ → $CO_{2}$ + $H_{2}O$

Firstly balance the following equation:

Before balancing the equation:

$CH_{4} + O_{2}$ → $CO_{2} + H_{2} O$

After balancing the equation:

$CH_{4} + 2O_{2}$ → $CO_{2} + 2 H_{2} O$

where;

$CH_{4}$ represents methane molecule

$O_{2}$ represents oxygen molecule

$CO_{2}$ represents carbon dioxide molecule

$H_{2}O$ represents water molecule

$CH_{4}$ +2 $O_{2}$ → $CO_{2}$ + 2 $H_{2}O$

16 grams of methane → 44 grams of carbon dioxide

6 × 10^-3 grams of methane → ?

= $\frac{44*6.00*10^{-3} }{16}$

= 16.5 × 10^-3 grams of carbon dioxide is produced from the complete combustion.

<u><em>Therefore 16.5 × 10^-3 grams of carbon dioxide is produced from the complete combustion.</em></u>

5 0

3 years ago