What is the oxidation number of chlorine

A water treatment tablet contains 20.0 mg of tetraglycine hydroperiodide, 40.0% of which is available as soluble iodine. If two

Galina-37 [17]

Answer:

$16\ \text{ppm}$

Explanation:

Mass of one tablet = 20 mg

Mass of two tablets = $2\times 20=40\ \text{mg}$

Percent that is soluble in water = 40%

Mass of tablet that is soluble in water = $0.4\times 40=16\ \text{mg}$

So, mass of solute is $16\ \text{mg}$

Density of water = 1 kg/L

Volume of water = 1 L

So, mass of 1 L of water is $1\times 1=1\ \text{kg}=1000\ \text{g}$

PPM is given by

$\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}\times 10^6=\dfrac{16\times 10^{-3}}{1000}\times 10^6\\ =16\ \text{ppm}$

Hence, the concentration of iodine in the treated water $16\ \text{ppm}$ .

8 0

2 years ago

Can swords shatter from cold temperatures

sp2606 [1]

Depends on how the sword is made, what materials are used and temperature used but yes they can shatter.

When molecules cool down they stop vibrating and moving as much and so they "shrink" and the metal of the sword becomes brittle. sometimes they shrink at different phases which cause tension in the sword if this tension is strong enough it can cause the metallic bonds to break causing the sword to shatter.

hope that helps

6 0

3 years ago

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibriu

tatyana61 [14]

Answer:

[H2] = 0.012 M

[N2] = 0.019 M

[H2O] = 0.057 M

Explanation:

The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.

2NO(g) + 2H2(g) ⇒ N2(g) + 2H2O(g)

i mol 0.10 0.050 0.10

c mol -0.038 -0.038 +0019 +0.038

e mol 0.062 0.012 00.019 0.057

Since the volume of the vessel is 1.0 L, the concentrations in molarity are:

[NO] = 0.062 M

[H2] = 0.012 M

[N2] = 0.019 M

[H2O] = 0.057 M

5 0

2 years ago

In a room full of air, the air is mainly composed of Nitrogen and Oxygen molecules (both at room temperature). Find (to two sign

zloy xaker [14]

Explanation:

Expression for the $v_{rms}$ speed is as follows.

$v_{rms} = \sqrt{\frac{3kT}{M}}$

where, $v_{rms}$ = root mean square speed

k = Boltzmann constant

T = temperature

M = molecular mass

As the molecular weight of oxygen is 0.031 kg/mol and R = 8.314 J/mol K. Hence, we will calculate the value of $v_{rms}$ as follows.

$v_{rms} = \sqrt{\frac{3kT}{M}}$

= $\sqrt{\frac{3 \times 8.314 J/mol K \times 309.02 K}{0.031 kg/mol}}$

= 498.5 m/s

Hence, $v_{rms}$ for oxygen atom is 498.5 m/s.

For nitrogen atom, the molecular weight is 0.028 kg/mol. Hence, we will calculate its $v_{rms}$ speed as follows.

$v_{rms} = \sqrt{\frac{3kT}{M}}$

= $\sqrt{\frac{3 \times 8.314 J/mol K \times 309.92 K}{0.028 kg/mol}}$

= 524.5 m/s

Therefore, $v_{rms}$ speed for nitrogen is 524.5 m/s.

3 0

3 years ago

PLEASE HELP!!! NEED TO PASS THIS TO THE FIRST SEMESTER!

Cloud [144]

Answer:

Reaction 1 is balanced but 2 is not balanced , the balance equation are :

1. $CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)$

2. $CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + H_{2}O(aq)$

Explanation:

Balanced Equations : These are the equation which follows the law of conservation of mass .

The total number of atoms present in reactant is equal to total number of atoms present in product.

1. $CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)$

This is acid - base type reaction where

$CH_{3}COOH(aq)$ act as Acid

$NaHCO_{3}(aq)$ act as weak base

Reactant : $CH_{3}COOH(aq)$ , $NaHCO_{3}$

Number of atoms of :

C = 2 ( $CH_{3}COOH(aq)$ ) + 1 ( $NaHCO_{3}$ )

= 2 + 1

= 3

H = 4( $CH_{3}COOH(aq)$ ) + 1 ( $NaHCO_{3}$ )

= 4 + 1

5

O = 2( $CH_{3}COOH(aq)$ ) + 3 ( $NaHCO_{3}$ )

= 5

Na = 1 ( $NaHCO_{3}$ )

= 1

Product : $CO_{2}(g)$ , $H_{2}O(l)$ , $CH_{3}COONa(aq)$

Number of atoms :

C = 1( $CO_{2}(g)$ ) + 2( $CH_{3}COONa(aq)$ )

= 1 + 2

= 3

H = 2( $H_{2}O(l)$ ) + 3( $CH_{3}COONa(aq)$ )

= 2 + 3

= 5

O = 1( $H_{2}O(l)$ ) + 2( $CH_{3}COONa(aq)$ )

+2( $CO_{2}(g)$

= 1 + 2 + 2

= 5

Na = 1( $CH_{3}COONa(aq)$

= 1

Number of Na =1 , C = 3 , H= 5 and O =5 in both reactant and product , so it is a balanced reaction

2. $CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + CO_{2}(g)$

This is double displacement reaction .

Check the balancing in both reactant and products should be :

Na = 2

H = 2

Ca = 1

C = 2

O = 6

Cl = 2

7 0

2 years ago