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3 years ago

Identify two factors that can make climates vary from place to place

Chemistry

2 answers:

Alekssandra [29.7K]3 years ago

8 0

Altitude and latitude.

Biomes of higher altitude are colder.
Biomes that are closer to the equator are warmer.

QveST [7]3 years ago

7 0

Altitude (height) and if the location is close to water

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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

4 0

2 years ago

1. Land-dwelling creatures depend on the the water cycle to ___________ fresh water from its salt solution in the Earth's oceans

Novosadov [1.4K]

1 is evaporate and 2 is precipitate

3 0

2 years ago

Read 2 more answers

2. In Experiment SOL, you investigated the solubility of oxalic acid. Sodium oxalate, Na2C2O¬4, is the sodium salt of this acid.

AnnZ [28]

Answer:

Sodium oxalate is a basic salt. In water it can be dissolved and dissociated.

The oxalic acid in water has two dissociations.

Explanation:

Na2C2O4 ---> 2Na+ + C2O4-2

Sodium oxalate is the conjugate base of a weak acid. In water this salt, dissociates completely giving rise to the sodium and oxalate ions. As Na+ comes from a strong base, in water it does not produce hydrolysis while oxalate does react in water, because it takes a proton from it and it generates a basic hydrolysis releasing OH-.

C2O4-2 + H2O ⇄ HC2O4- + OH-

In water the salt is basic. The pH of an aqueous solution of this salt is basic, since OH- is generated.

The HC2O4- has a second hydrolisis, it takes another proton from water to form oxalic acid.

HC2O4- + H2O ⇄ H2C2O4 + OH-

The oxalic acid acts as a weak acid, it can release 2 protons to water, to make oxalate (its conjugate base).

H2C2O4 + H2O ⇄ H3O+ + HC2O4-

HC2O4- + H2O ⇄ H3O+ C2O4-2

The HC2O4- acts as an ampholyte since it accepts and delivers protons simultaneously.

6 0

3 years ago

Which of the following substances found in tobacco smoke stimulates the brain? A. acetone B. toluene C. nicotine D. ammonia Plea

Vaselesa [24]

The answer is C. Nicotine is the substance found in tobacco smoke that stimulates the brain.

7 0

3 years ago

Read 2 more answers

Identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent for the following equation

bazaltina [42]

The element that gains electrons, becomes reduced.

While the one which loses electrons, becomes oxidized.

In this equation,

CH₃OH + Cr₂O₇²⁻---- --> CH₂O + Cr³⁺.

By balancing the equation, we will get:

3CH₃OH + Cr₂O₇²⁻ + 8H⁺ --> 3CH₂O + 2Cr³⁺ + 7H₂O

Here the oxidation state of Cr changes from +6 to +3 that is it is being reduced thus serving as a oxidizing agent while other element retain their charges.

Here Cr₂O₇²⁻ is reduced while CH₃OH is oxidized.

So Cr₂O₇²⁻ serves as a oxidizing agent, while CH₃OH serves as reducing agent .

4 0

2 years ago