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2 years ago

In the presence of a base, blue litmus paper will .O turn purplestay blueto turn red

Chemistry

2 answers:

Inessa05 [86]2 years ago

7 0

Answer:

In the presence of a base, blue litmus paper will turn red........

Strike441 [17]2 years ago

5 0

Answer:

In the presence of a base the blue litmus paper would remain blue.

Explanation:

This is because the litmus paper will react depending on the pH of the solution added. If the solution is acidic pH < 7, the litmus paper will turn red. If the solution is basic, pH > 7 the litmus paper will turn blue. If the litmus paper is blue when a base is added it will remain blue.

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A block of aluminum occupies a volume of 15.0 mL and weighs 45 grams.

vladimir2022 [97]

Answer: density equals 3 g/mL

Step by step explanation:

D=m/v
D=45/15
D=3

5 0

3 years ago

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]

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

Explanation:

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

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Answer:

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Explanation:

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2 years ago

What is the total number of atoms in this formula? NaHCO3

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6 - one sodium atom, 1 hydrogen atom, 1 carbon atom, and 3 oxygen atoms.

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3 years ago

How did the pH level and the water components level change after adding water to the battery acid?

ololo11 [35]

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3 years ago

In the presence of a base, blue litmus paper will .O turn purplestay blueto turn red​

In the presence of a base, blue litmus paper will .O turn purplestay blueto turn red