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3 years ago

13

Which is the most common type of precipitation

3 answers:

aliina [53]3 years ago

5 0

Rain is the most common type of precipitation. The conditions for the other types are more rare than those required for rain.

zhannawk [14.2K]3 years ago

5 0

Answer:

The Answer is Rain

Explanation:

Just did It it is correct

salo2 years ago

0 0

RRAAIINN
RAIN :)

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The building block of all matter is a) water b) oxygen c) an atom d) an element

jeka94

My best try would be d an element

4 0

3 years ago

Calculate the molarity of an HCl solution if 23.88 mL of it reacts with 6.5287 grams of sodium carbonate (106 g/mol) according t

Ivanshal [37]

Answer:

5.158 mol/L

Explanation:

To find the molarity, you need to use the formula:

Molarity (M) = moles / volume (L)

You have been grams sodium carbonate. You need to (1) convert grams Na₂CO₃ to moles (via molar mass), then (2) convert moles Na₂CO₃ to moles HCl (via mole-to-mole ratio from equation), then (3) convert mL to L (by dividing by 1,000), and then (4) use the molarity equation.

<u>Steps 1 - 2:</u>

2 HCl + 1 Na₂CO₃ ----> 2 NaCl + H₂O + CO₂

6.5287 g Na₂CO₃ 1 mole 2 moles HCl
-------------------------- x ------------- x ------------------------- = 0.12318 mole HCl
106 g 1 mole Na₂CO₃

<u>Step 3:</u>

23.88 mL / 1,000 = 0.02388 L

<u>Step 4:</u>

Molarity = moles / volume

Molarity = 0.12318 mole / 0.02388 L

Molarity = 5.158 mole/L

**mole/L is equal to M**

7 0

2 years ago

Hard water contains relatively large concentrations of

ycow [4]

Answer:

<u><em>D.</em></u>

<u><em>All of these ions</em></u>

Explanation:

Hope this helps:)

5 0

3 years ago

Read 2 more answers

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

3 years ago

Uses of carbon dioxide.

andrey2020 [161]

Answer:

hello can you follow me

Explanation:

used in fire extinction, blasting rubber, foaming rubber and plastic

6 0

3 years ago