Which are different structures of the eye? Select four options.

What is the oxidation number of cl− in the hypochlorite ion clo−?

mariarad [96]

Answer : The oxidation number of chlorine (Cl) is, (+1)

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of oxygen (O) in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

The given compound is, $ClO^-$

Let the oxidation state of 'Cl' be, 'x'

$x+(-2)=-1\\\\x-2=-1\\\\x=+1$

Therefore, the oxidation number of chlorine (Cl) is, (+1)

5 0

3 years ago

HELP WITH FINAL CHEM

atroni [7]

Answer:

idvd has a very high performance and a lot more than just the right side to get it all out

3 0

3 years ago

Predict the boiling point of water at a pressure of 1.5 atm.

Lina20 [59]

Answer:

100.8 °C

Explanation:

The Clausius-clapeyron equation is:

$ln\frac{P_{1} }{P_{2}} =$ -Δ $\frac{H_{vap}}{r} (\frac{1}{T_{2}}-\frac{1}{T_{1}} )$

Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'

Isolating for T2 gives:

$T_{2}=(\frac{1}{T_{1}} -\frac{Rln\frac{P_{2}}{P_{1}} }{Delta H_{vap}}$

(sorry for 'deltaHvap' I can not input symbols into equations)

thus T2=100.8 °C

7 0

3 years ago

Choose the correct products for the double replacement reaction below. Click here to access the solubility rules to determine wh

Lemur [1.5K]

MgCl₂ & PbSO₄ is the correct product for the given double replacement reaction.

<h3>What is double displacement reaction?</h3>

In the double displacement reaction displacement of two substrate will take place among two reactants and formation of products takes place.

In the question, PbCl₂ reacts with MgSO₄ and displacement between the cations takes place and formation of MgCl₂ and insoluble precipitate PbSO₄ is formed.

Given reaction is represented as:

PbCl₂ + MgSO₄ → MgCl₂ + PbSO₄

According to the activity series, reactivity of Magnesium is more as compared to the lead atom. So lead is displaced by the magnesium atom and form the above given products.

Hence, option (3) is correct i.e. MgCl₂ & PbSO₄.

To know more about double displacement reaction, visit the below link:

brainly.com/question/26413416

8 0

2 years ago

Read 2 more answers

A chemist must prepare of sodium hydroxide solution with a pH of at . He will do this in three steps: Fill a volumetric flask ab

77julia77 [94]

Answer:

0.0400 g for the example given below.

Explanation:

pH value is not provided, so we'll solve this problem in a general case and then we will use an example to justify it.

By definition, $pH = -log[H_3O^+]$ .
NaOH is a strong base, as it's a hydroxide formed with a group 1A metal, so it dissociates fully in water by the equation: $NaOH (aq)\rightarrow Na^+ (aq) + OH^- (aq)$ .
From the equation above, using stoichiometry we can tell that the molarity of hydroxide is equal to the molarity of NaOH: $[NaOH] = [OH^-]$ .
Concentration of hydroxide is then equal to the ratio of moles of NaOH and the volume of the given solution. Moles themselves are equal to mass over molar mass, so we obtain: $[OH^-] = [NaOH] = \frac{n_{NaOH}}{V} = \frac{m_{NaOH}}{M_{NaOH}V}$ .
We also know that $pOH = 14.00 - pH = -log[NaOH]$ . Take the antilog of both sides: $10^{-pOH} = 10^{pH - 14.00} = [NaOH] = \frac{m_{NaOH}}{M_{NaOH}V}$ .
Solve for the mass of NaOH: $m_{NaOH} = 10^{pH - 14.00}\cdot M_{NaOH}\cdot V$ .

Now, let's say that pH is given as 12.00 and we use a 100-ml volumetric flask. Then we would obtain:

$m_{NaOH} = 10^{12.00 - 14.00}\cdot 39.997 g/mol\cdot 0.100 L = 0.0400 g$

7 0

3 years ago