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2 years ago

Will sodium iodide react with bromine to produce sodium bromide and iodine? Why or why not?

Chemistry

1 answer:

Gre4nikov [31]2 years ago

4 0

Answer:

Explanation:

The higher the period the higher the activity of an element, therefore, since iodine is in period 6 and bromine is in period 5, the described reaction is not possible due to the fact that bromine is less active

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The cylinder of aluminum in class was approximately 13mm in circumference and 42mm in length. What would be it's approximate mas

Roman55 [17]

Answer:

Mass, m = 1.51 grams

Explanation:

It is given that,

The circumference of Aluminium cylinder, C = 13 mm = 1.3 cm

Length of the cylinder, h = 4.2 cm

We know that the density of the Aluminium is 2.7 g/cm³

Circumference, C = 2πr

$r=\dfrac{C}{2\pi}\\\\r=\dfrac{1.3}{2\pi}\\\\r=0.206\ cm$

Density is equal to mass per unit volume.

$d=\dfrac{m}{V}$

m is mass of the cylinder

V is the volume of the cylinder

$V=\pi r^2h\\\\V=\dfrac{22}{7}\times0.206^2\times 4.2\\\\V=0.5601\ cm^3$

So,

$m=d\times V\\\\m=2.7\times 0.5601\\\\m=1.51\ g$

So, the mass of the cylinder is 1.51 grams.

5 0

4 years ago

The gold foil experiment resulted in two major discoveries which were made about the structure of the atom. what were they

DanielleElmas [232]

It showed that the nucleus is positively charged, and that the atom isn't just made of electrons.

8 0

3 years ago

A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base

tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ = 0.00375 mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

$HN_3 + OH- ---> N^-_{3} + H_2O$

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L

Concentration of $HN_3$ = $\dfrac{0.001755}{0.0383}$ = 0.0458 M

Concentration of $N^{-}_3$ = $\dfrac{ 0.001995 }{0.0383}$ = 0.0521 M

GIven that :

Ka = $1.9 x 10^{-5}$

Thus; it's pKa = 4.72

$pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})$

$pH =4.72 + log(1.1376)$

$pH =4.72 + 0.05598$

$pH =4.77598$

pH ≅ 4.80

3 0

3 years ago

Li + HNO3 > LiNO3 + H2 how do I balance it? Show work pls

azamat

Answer: The balanced equation is $2Li + 2HNO_{3} \rightarrow 2LiNO_{3} + H_{2}$ .

Explanation:

The given reaction equation is as follows.

$Li + HNO_{3} \rightarrow LiNO_{3} + H_{2}$

Number of atoms present on reactant side are as follows.

Li = 1
H = 1
$NO_{3}$ = 1

Number of atoms present on product side are as follows.

Li = 1
H = 2
$NO_{3}$ = 1

To balance this equation, multiply Li by 2 and $HNO_{3}$ by 2 on reactant side. Also, multiply $LiNO_{3}$ by 2 on product side.

Hence, the equation can be rewritten as follows.

$2Li + 2HNO_{3} \rightarrow 2LiNO_{3} + H_{2}$

Now, number of atoms present on reactant side are as follows.

Li = 2
H = 2
$NO_{3}$ = 2

Number of atoms present on product side are as follows.

Li = 2
H = 2
$NO_{3}$ = 2

As there are same number of atoms on both reactant and product side. Hence, the equation is now balanced.

Thus, we can conclude that the balanced equation is $2Li + 2HNO_{3} \rightarrow 2LiNO_{3} + H_{2}$ .

3 0

3 years ago

An old iron battery was discarded in this meadow. The iron in the battery is undergoing a ____ change as it rusts. The rusting i

Lelu [443]

Answer:

chemical

Explanation:

A chemical change is irreversible and leads to the formation of new products.

Rusting occurs as a result of an irreversible chemical reaction between iron, oxygen and moisture. Rusting is the oxidation of iron when exposed to oxygen and moisture.

The rusting of iron is an electrochemical process in which water serves as the electrolyte and iron serves as the anode. The rusting of iron leads to the formation of hydrated iron III oxide (rust) usually seen an brown flakes that surround a rusted iron material.

3 0

3 years ago