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Pavlova-9 [17]
3 years ago
14

Which substance is a binary acid

Chemistry
1 answer:
Bumek [7]3 years ago
4 0
<span>It is known that acids compounds contains hydrogen and produces hydrogen ion in water. A binary acid however is an acid that have two elements, one of the element has a hydrogen attached to it. Examples of binary acids are hydrogen fluoride (HF), hydrogen bromide (HBr) and hydrogen sulfide (H2S). In naming a binary acid, it has two rules; one, as pure compounds and two, as acid solutions. For pure compounds, start with the name ‘hydrogen’ and end the anion name with ‘-ide’. For acidic compounds, start with ‘hydro-‘, end the anion with ‘-ic’ and add ‘acid’.</span>
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Many medical devices are designed using the electromagnetic spectrum. Which type of electromagnetic wave has enough energy to ki
tresset_1 [31]
Gamma rays because the kill every living cell and they have the shortest waves that are dangerous
4 0
3 years ago
Is the law of conservative mass observed in this equation CaCO3 + 2HCI --&gt;CaCI2 +H2O + CO2
pychu [463]

Answer:

The law is observed in the given equation.

Explanation:

CaCO₃ + 2HCI → CaCI₂ +H₂O + CO₂

In order to find out if the law of conservative mass is followed, we need to <u>count how many atoms of each element are there in both sides of the equation</u>:

  • Ca ⇒ 1 on the left, 1 on the right.
  • C ⇒ 1 on the left, 1 on the right.
  • O ⇒ 3 on the left, 3 on the right.
  • H ⇒ 2 on the left, 2 on the right.
  • Cl ⇒ 2 on the left, 2 on the right.

As the numbers for all elements involved are the same, the law is observed in the given equation.

8 0
3 years ago
In each row check off the boxes that apply to the highlighted reactant. reaction The highlighted reactant acts as a... (check al
tekilochka [14]

The given question is incomplete. The complete question is :

In each row check off the boxes that apply to the underlined reactant. The underlined reactant acts as a... (check all that apply)

1. HCH_3CO_2(aq)+NH_3(aq)\rightarrow CH_3COO^-(aq)+NH_4^+(aq)

here underlined is HCH_3CO_2

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

2. BH_3(aq)+NH_3(aq)\rightarrow BH_3NH_3(aq)

Here underlined is NH_3

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

3. HNO_2(aq)+C_2H_5NH_2(aq)\rightarrow NO_2^-(aq) + C_2H_5NH_3^+(aq)

Here underlined is C_2H_5NH_2

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

Answer: 1. Brønsted-Lowry acid

2. Lewis base

3. Brønsted-Lowry base

Explanation:

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

1.  HCH_3CO_2(aq)+NH_3(aq)\rightarrow CH_3CO^{2-}(aq)+NH_4^+aq)

As HCH_3CO_2(aq) is donating a proton , it acts as a bronsted acid.

2. BH_3(aq)+NH_3(aq)\rightarrow BH_3NH_3(aq)

As NH_3 contains a lone pair of electron on nitrogen , it can easily donate electrons to BH_3 and act as lewi base.

3.  HNO_2(aq)+C_2H_5NH_2(aq)\rightarrow NO_2^-(aq) + C_2H_5NH_3^+(aq)

As C_2H_5NH_2(aq) is accepting a proton , it acts as a bronsted base.

7 0
3 years ago
The value of the solubility product constant for Ag2CO3 is 8.5 × 10‒12 and that of Ag2CrO4 is 1.1 × 10‒12. From this data, what
Lena [83]

Answer:

B) 7.7

Explanation:

For the reaction    Ag2CO3(s) + CrO42‒(aq) → Ag2CrO4(s) + CO32‒(aq)

Kc = (CO₃²⁻) / (CrO₄²⁻)

and the Ksp given are

Ag₂CO₃    ⇒  2 Ag⁺(aq) + CO₃²⁻(aq)    Ksp₁ = (Ag⁺)²(CO₃²⁻)  

Ag₂CrO₄   ⇒  2 Ag⁺(aq)+ CrO₄²⁻(aq)   Ksp₂ = (Ag⁺)²(CrO₄²⁻)

Where (...) indicate concentrations M

Notice if we divide the expressions for Ksp we get:

Ksp₁/Ksp₂ = (CO₃²⁻)  / (CrO₄²⁻) = 8.5 x 10⁻¹² / 1.1 x 10⁻¹² = 7.7

which is the desired answer.

7 0
3 years ago
A photon of light possesses 5 x 10^-19 J of energy. Calculate its frequency
saveliy_v [14]

Answer:

The frequency of photon is 0.75×10¹⁵ s⁻¹.

Explanation:

Given data:

Energy of photon = 5×10⁻¹⁹ J

Frequency of photon = ?

Solution:

Formula;

E = hf

h = planck's constant = 6.63×10⁻³⁴ Js

5×10⁻¹⁹ J =  6.63×10⁻³⁴ Js ×f

f =  5×10⁻¹⁹ J / 6.63×10⁻³⁴ Js

f = 0.75×10¹⁵ s⁻¹

The frequency of photon is 0.75×10¹⁵ s⁻¹.

4 0
3 years ago
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