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3 years ago

Using the phase diagram for H2O, which of the following correctly describes water at 0°C and 1 atm?

Chemistry

2 answers:

SCORPION-xisa [38]3 years ago

6 0

At the melting point. Draw a line up from 0 degrees and a line to the right from 1 atm. They meet at the line between solid and liquid... the melting point

7nadin3 [17]3 years ago

3 0

just took the quiz and its

at the boiling point

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What is the mass of 0.572 moles of Al?

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Answer: 15.433704

Use the periodic table to check the atomic mass, this is the number of grams per mole → 1 mole of Aluminum is 26.982 g

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3 years ago

A beaker of nitric acid is neutralized with dilute aqueous solution of calcium hydroxide.Write a net ionic equation for this rea

Soloha48 [4]

Complete balanced equation: 2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O

Ionized equation (with spectator ions):
2H⁺ + 2NO₃⁻ + Ca²⁺ + 2OH⁻   → Ca²⁺ + 2NO₃⁻ + 2H₂O

By eliminating the ions that are the same of both sides of the equation (spectator ions):

2H⁺ + 2OH⁻   → 2H₂O [Net Ionic Equation]

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3 years ago

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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3 years ago

Part iv. Is the neutralization reaction enthalpy favored?

Burka [1]

Yes, it is a special case of enthalpy of neutralization.

The enthalpy of neutralization (ΔHn) is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt.

The standard enthalpy change of neutralization is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water.

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3 years ago

Splitting nuclei is called __________ and joining nuclei is called _________.

SCORPION-xisa [38]

The answer is B, nic fiss, nic fusi

8 0

3 years ago

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