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Scorpion4ik [409]
3 years ago
15

Is HNO3 an acid or a baseI NEED HELP ASAP​

Chemistry
2 answers:
katrin [286]3 years ago
6 0

Answer:

It is a strong acid

Explanation:

Artyom0805 [142]3 years ago
3 0

Answer:

HNO3 is a potent acid, a base, a nitrating agent and a heavy oxidising agent at times. In the presence of a stronger acid, it serves as a base.

Explanation:

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How many electrons, protons, neutrons, and what is the atomic number and atomic mass (if any) does Fluoride have?
ExtremeBDS [4]
Fluoride is an anion of Fluorine

What this means is that the two have the same number of protons (9), but Fluoride has 10 electrons compared to Fluorine's 9.

So the answers are:
Protons - 9
Neutrons - 9
Electrons - 10
Atomic Number - 9
Atomic Mass - 19 g/mol
5 0
3 years ago
Read 2 more answers
The metal content of iron in ores can be determined by a redox procedure in which the sample is first oxidized with Br2 to conve
coldgirl [10]

Answer:

80.27%

Explanation:

Let's consider the following balanced equation.

2 Fe³⁺(aq) + Sn²⁺(aq) ⇒ 2Fe²⁺(aq) + Sn⁴⁺(aq)

First, we have to calculate the moles of Sn²⁺ that react.

\frac{0.1015molSn^{2+} }{1L} .13.28 \times 10^{-3} L=1.348\times 10^{-3}molSn^{2+}

We also know the following relations:

  • According to the balanced equation, 1 mole of Sn²⁺ reacts with 2 moles of Fe³⁺.
  • 1 mole of Fe³⁺ is oxidized from 1 mole of Fe.
  • The molar mass of Fe is 55.84 g/mol.

Then, for 1.348 × 10⁻3 moles of Sn²⁺:

1.348\times 10^{-3}molSn^{2+}.\frac{2molFe^{3+} }{1molSn^{2+} } .\frac{1molFe}{1molFe^{3+} } .\frac{55.84gFe}{1molFe} =0.1505gFe

If there are 0.1505 g of Fe in a 0.1875 g sample, the mass percentage of Fe is:

\frac{0.1505g}{0.1875g} \times 100 \% = 80.27\%

5 0
3 years ago
Calculate the area of a 3.0 inch by 5.0 inch index card in square millimeters (mm). (You can look up the formula for the area of
meriva

Answer:

The area of the given rectangular index card = <u>9677.4 mm²</u>    

Explanation:

Area is defined as the space occupied by a two dimensional shape or object. The SI unit of area is square metre (m²).

<u>The area of a rectangle</u> (A) =  length (l) × width (w)

Given dimensions of the rectangle: Length (l) = 5.0 inch, Width (w) = 3.0 inch

Since, 1 inch = 25.4 millimetres (mm)

Therefore, l = 5 × 25.4 = 127 mm, and w = 3 × 25.4 = 76.2 mm

Therefore, <u>the area of the given rectangular index card</u> = A= l × w = 127 mm × 76.2 mm = <u>9677.4 mm²</u>

5 0
3 years ago
What amount (moles) is represented by each of these samples?
Zinaida [17]
<h3>Answer:</h3>

                  a)  Moles of Caffeine  =  1.0 × 10⁻⁴ mol

                  b) Moles of Ethanol   =  4.5 × 10⁻³ mol

<h3>Solution:</h3>

Data Given:

                  Mass of Caffeine  =  20 mg  =  0.02 g

                  M.Mass of Caffeine  =  194.19 g.mol⁻¹

                  Molecules of Ethanol  =  2.72 × 10²¹

Calculate Moles of Caffeine as,

                               Moles  =  Mass ÷ M.Mass

Putting values,

                               Moles  =  0.02 g ÷ 194.19 g.mol⁻¹

                                Moles  =  1.0 × 10⁻⁴ mol

Calculate Moles of Ethanol as,

                                                         As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Particles and Avogadro's Number is given as,

                          Number of Moles  =  Number of Molecules ÷ 6.022 × 10²³

Putting values,

                          Number of Moles  =  2.72 × 10²¹ Molecules ÷ 6.022 × 10²³

                          Number of Moles  =  4.5 × 10⁻³ Moles

5 0
3 years ago
What is the pH of a 0.300 M NH₃ solution that has Kb = 1.8 × 10⁻⁵ ? The equation for the dissociation of NH₃ is: NH₃ (aq) + H₂O
nalin [4]

Answer:

11.4

Explanation:

Step 1: Given data

  • Concentration of the base (Cb): 0.300 M
  • Basic dissociation constant (Kb): 1.8 × 10⁻⁵

Step 2: Write the dissociation equation

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

Step 3: Calculate the concentration of OH⁻

We will use the following expression.

[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8  \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M

Step 4: Calculate the pOH

We will use the following expression.

pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6

Step 5: Calculate the pH

We will use the following expression.

pH+pOH=14\\pH = 14-pOH = 14-2.6 = 11.4

8 0
3 years ago
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