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3 years ago

How much 0.2M HCl can be made from 5.0mL of a 12.0M HClSolution?

Chemistry

1 answer:

USPshnik [31]3 years ago

7 0

Answer: The volume of 0.10 M NaOH required to neutralize 30 ml of 0.10 M HCl is, 30 ml.

Explanation:

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock $HCl$ solution = 12.0 M

$V_1$ = volume of stock $HCl$ solution = 5.0 ml

$M_2$ = molarity of dilute $HCl$ solution = 0.2 M

$V_2$ = volume of dilute $HCl$ solution = ?

Putting in the values we get:

$(12.0M)\times 5.0=0.2\times V_2$

$V_2=300ml$

Therefore, 300 ml of 0.2M HCl can be made from 5.0mL of a 12.0M $Hcl$ solution.

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What is the mass in grams for 7.14x10^3 mol of PCI3?

Shalnov [3]

Answer:

980,536.2 grams

Explanation:

The mass of a compound is equal to its moles times its molar mass.

The molar mass of PCl3 is 137.33 g/mol

137.33 x 7.14 x 10^3 = 980,536.2

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2 years ago

When baking soda (sodium bicarbonate or sodium hydrogen carbonate, NaHCO3) is heated, it releases carbon dioxide gas, which is r

MariettaO [177]

Answer:

a. NaHCO₃ + HCl → NaCl + H₂O + CO₂

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A possible reaction for NaHCO₃ to make dioxide is this one, when it reacts with hydrochloric to produce the mentioned gas.

NaHCO₃ + HCl → NaCl + H₂O + CO₂

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So 1 mol of baking soda, produce 1 mol of CO₂

Let's calculate the moles

20.5 g CO₂ / 44 g/m = 0.466 moles

This moles of gas came from the same moles of salt.

Molar mass baking soda = 84 g/m

Molar mass . moles = mass

84 g/m . 0.466 moles = 39.14 g

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4 years ago

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Answer:

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3 years ago

How did Earth come to have an oxygen atmosphere?A.Precambrian rocks released oxygen into the atmosphere.B.Volcanoes released oxy

Luba_88 [7]

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4 years ago

Give the name and formula of the compound formed from the following elements: a) Sodium and nitrogen b) Oxygen and strontium c)

elena-s [515]

Answer: a) Sodium and nitrogen : $NaN_3$ : sodium nitride.

b) Oxygen and strontium : $SrO$ : strontium oxide.

c) Aluminum and chlorine : $AlCl_3$ : aluminium chloride.

d) Cesium and bromine : $CsBr$ : cesium bromide.

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

a) Sodium and nitrogen : Here sodium is having an oxidation state of +1 called as $Na^{+}$ cation and nitrogen forms an anion $N^{3-}$ with oxidation state of -3. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $NaN_3$ named as sodium nitride.

b) Oxygen and strontium : Here strontium is having an oxidation state of +2 called as $Sr^{+}$ cation and oxygen forms an anion $O^{2-}$ with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $SrO$ named as strontium oxide.

c) Aluminum and chlorine : Here aluminium is having an oxidation state of +3 called as $Al^{3+}$ cation and chlorine forms an anion $Cl^{-}$ with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $AlCl_3$ named as aluminium chloride.

d) Cesium and bromine : Here cesium is having an oxidation state of +1 called as $Cs^{+}$ cation and bromine forms an anion $Br^{-}$ with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $CsBr$ named as cesium bromide.

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3 years ago