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2 years ago

Determine whether each chemical substance would remain the same color or turn pink in the presence of phenolphthalein.

Chemistry

1 answer:

Serhud [2]2 years ago

3 0

Answer:

See the answer below

Explanation:

<em>The complete question can be seen in the attached image.</em>

<u>Phenolphthalein is an indicator that is often utilized in an acid-base reaction to indicate the endpoints of such reactions due to its ability to change color from pink/colorless to colorless/pink depending on if the final solution is acidic or basic.</u>

Phenolphthalein is usually colorless in acidic solutions and appears pink in basic solutions. The more basic or alkaline a solution is, the stronger the pink color of phenolphthalein. Hence;

1. Ammonia with a pH of 11 is basic, phenolphthalein will turn pink.

2. Battery acid with a pH of 1 is acidic, it will remain colorless.

3. Lime juice with a pH of 2 is acidic, it will remain colorless.

4. Mashed avocado with a pH of 6.5 is acidic, it will remain colorless.

5. Seawater with a pH of 8.5 is basic, it will turn pink.

6. Tap water with a pH of 7 is neutral, it will remain colorless

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Dr. Franck cuts a bar of pure gold into smaller and smaller pieces. Will this action change the element that makes up the bar? E

Anvisha [2.4K]

Answer:

i believe this is a chemical or physical question? well your answer to that is no the element does not change because the gold is still gold it is still physical because you have just cut it into piece it is still gold

Explanation:

lmk if it was helpful :/

7 0

1 year ago

Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic

Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9$

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = $164.5\times 10^{-9} g$

$1 ng=10^{-9} g$

Volume of the sample = V = $15.3 cm^3$

Density of the lake water sample ,d= $1.00 g/cm^3$

Mass of sample = M = $d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g$

$ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75$

10.75 is the concentration of arsenic in the sample in parts per billion.

Mass of arsenic in $1 cm^3$ of lake water = $\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g$

Mass of arsenic in $0.710 km^3$ lake water be m.

$1 km^3=10^{15} cm^3$

Mass of arsenic in $0.710\times 10^{15} cm^3$ lake water :

$m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g$

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

It will take 1.37 years to remove all of the arsenic from the lake.

3 0

1 year ago

How many grams of NO are required to produce 145 g of N2 in the following reaction?

V125BC [204]

Answer:

b. 186 g

Explanation:

Step 1: Write the balanced equation.

4 NH₃(g) + 6 NO(g) → 5 N₂(g) + 6 H₂O(l)

Step 2: Calculate the moles corresponding to 145 g of N₂

The molar mass of nitrogen is 28.01 g/mol.

$145g \times \frac{1mol}{28.01 g} =5.18 mol$

Step 3: Calculate the moles of NO required to produce 5.18 moles of N₂

The molar ratio of NO to N₂ is 6:5.

$5.18molN_2 \times \frac{6molNO}{5molN_2} = 6.22molNO$

Step 4: Calculate the mass corresponding to 6.22 moles of NO

The molar mass of NO is 30.01 g/mol.

$6.22mol \times \frac{30.01g}{mol} =186 g$

4 0

1 year ago

What volume in milliliters of 6.0 M NaOH is needed to prepare 175mL of 0.20 M NaOH by dilution?

Ilya [14]

Answer:

V¹N²= V²N²

here V¹= ?

N¹= 6.00

V²= 175ml

M²= 0.2M

So V¹= (V²N²)/N² = (175 x 0.2)/6

V¹ = 5.83 ml

Explanation:

Therefore diluting 5.83 ml of 6.00M NaOH to 175 m l ,we get 0.2M Solution.

4 0

1 year ago

6) A 0.20 ml CO2 bubble in a cake batter is at 27°C. In the oven it gets

Nata [24]

Answer: The new volume of cake is 1.31 mL.

Explanation:

Given: $V_{1}$ = 0.20 mL, $T_{1} = 27^{o}C$

$V_{2}$ = ?, $T_{2} = 177^{o}C$

Formula used to calculate new volume is as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{0.20 mL}{27^{o}C} = \frac{V_{2}}{177^{o}C}\\V_{2} = 1.31 mL$

Thus, we can conclude that the new volume of cake is 1.31 mL.

5 0

1 year ago