An aqueous solution has a hydroxide-ion concentration of 1.0 x 10-3M. What is the pH of the solution?

If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?

ElenaW [278]

Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$ where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation: $T_C+273=T_K$

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, $T_1 = 273[K]$ , and $T_2 = (49.4+273)[K]=322.4[K]$ .

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}$

$\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})$

$1810.04571428[mL]=V_2$

Adjusting for significant figures, this gives $V_2=1810[mL]$

4 0

1 year ago

Write a complete, balanced chemical equation where tin metal reacts with aqueous hydrochloric acid to produce tin(II) chloride a

AleksAgata [21]

Answer:

1. The balanced equation is given below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

2a. H is oxidized.

2b. Sn is reduced.

Explanation:

1. Balanced equation for the reaction between tin (Sn) metal and aqueous hydrochloric acid (HCl) to produce tin(II) chloride (SnCl₂) and hydrogen gas (H₂).

This is illustrated below:

Sn (s) + HCl (aq) –> SnCl₂ (aq) + H₂ (g)

There are 2 atoms of Cl on the right side and 1 atom on the left side. It can be balance by putting 2 in front of HCl as shown below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

Now, the equation is balanced

2. Determination of the element that is oxidize and reduced.

This can be obtained as follow:

We shall determine the change in oxidation number of each element.

NOTE:

a. The oxidation number of H is always +1 except in hydrides where it is –1.

b. The oxidation state of Cl is always –1.

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

For Tin (Sn):

Sn = 0

SnCl₂ = 0

Sn + 2Cl = 0

Cl = – 1

Sn + 2(–1) = 0

Sn – 2 = 0

Collect like terms

Sn = 0 + 2

Sn = +2

Therefore, the oxidation number of Tin (Sn) changes from 0 to +2

For H:

H = +1

H₂ = 0

The oxidation number of H changes from +1 to 0

For Cl:

Cl is always –1. Therefore no change.

Summary:

Element >>Change in oxidation number

Sn >>>>>>>From 0 to +2

H >>>>>>>>From +1 to 0

Cl >>>>>>>No change

Therefore,

Sn is reduced since its oxidation number increased from 0 to +2.

H is oxidized since it oxidation number reduced from +1 to 0

4 0

2 years ago

Help fast !

jok3333 [9.3K]

Answer:

1 mole represents 6.023×1023 particles.

1 mole of iodine atom= 6.023×1023

Given 127.0g of iodine.

no. of iodine atom = 1 mole of iodine

1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg

Given 48g of Mg = 2×6.023×1023

no. of Mg = 2 moles of Mg

1 mole of chlorine atom= 6.023× 1023

no. of chlorine atom = 35.5g of chlorine atom

Given 71g of chlorine atom=2× 6.023× 1023

no. of chlorine atom = 6.023×1023

2 moles of chlorine atom.

Given that 4g of hydrogen atom.

will be equal to 4 × 6.023 × 1023

no. of atoms of hydrogen= 4 moles of hydrogen atom.

7 0

3 years ago

Which of the following is an example of rapid combustion?

rjkz [21]

phosphorus in water is rapid combustion.

6 0

2 years ago

Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 4.90 moles of magnesium perchlorate, Mg(ClO4)2.]\

Leviafan [203]

Answer:

4.90 moles of $Mg(ClO_4)_2$ will produce (9.8) moles of $Cl^{-}$ ,

(4.90) moles of $Mg^{2+}$ and

(39.2) moles of $O^{2-}$

Explanation:

From the question we are told that

The number of moles of is $n = 4.90 \ mols$

The formation reaction of $Mg(ClO_4)_2$ is

$Mg^{2+} + 2 Cl^{-} + 8O^{2+} \to Mg(ClO_4)_2$

From the reaction we see that

1 mole of $Mg(ClO_4)_2$ is formed by 2 moles of $Cl^{-}$ 1 mole of $Mg^{2+}$ and 4 $O^{2-}$

This implies that

4.90 moles of $Mg(ClO_4)_2$ will produce (2 * 4.90) moles of $Cl^{-}$ ,

(1 * 4.90) moles of $Mg^{2+}$ and

(8 * 4.90) moles of $O^{2-}$

So

4.90 moles of $Mg(ClO_4)_2$ will produce (9.8) moles of $Cl^{-}$ ,

(4.90) moles of $Mg^{2+}$ and

(39.2) moles of $O^{2-}$

4 0

3 years ago

Read 2 more answers