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3 years ago

Use complete sentences to differentiate between acids and bases on the basis of taste. Give an example of each type.

Chemistry

1 answer:

Dmitry_Shevchenko [17]3 years ago

8 0

Answer:

The general taste in acids is sour and example is seen in citric acid in citrus fruits), while bases tastes bitter or even soapy and example is seen in NaOH

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A paint chip is placed in ethanol and curls what type of paint is it

Licemer1 [7]

Answer:

it is probably acrylic or latex

Explanation:

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2 years ago

HELP ASAP! GIVING BRAINLIEST!!<br><br> Why would a pendulum stop swinging? Where did the energy go?

kvasek [131]

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The pendulum would stop swinging because the kinetic energy would slow down for the friction is causing it to slow. The energy would go into potential energy.

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3 years ago

Read 2 more answers

What volume of 0.500 m h2so4 is needed to react completely with 20.0 ml of 0.400 m lioh?

ExtremeBDS [4]

The balanced chemical equation for the above reaction is as follows;
2LiOH + H₂SO₄ ---> Li₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
Number of OH⁻ moles reacted = number of H⁺ moles reacted at neutralisation
Number of LiOH moles reacted = 0.400 M / 1000 mL/L x 20.0 mL = 0.008 mol
number of H₂SO₄ moles reacted - 0.008 mol /2 = 0.004 mol
Number of H₂SO₄ moles in 1 L - 0.500 M
This means that 0.500 mol in 1 L solution
Therefore 0.004 mol in - 1/0.500 x 0.004 = 0.008 L
therefore volume of acid required = 8 mL

4 0

3 years ago

If 45.0 mL of a 0.0500 M HNO3, 10.0 mL of a 0.0500 M KSCN, and 30.0 mL of a 0.0500 M Fe(NO3)3 are combined, what is the initial

jeka94

Answer:

the initial concentration of SCN- in the mixture is 0.00588 M

Explanation:

The computation of the initial concentration of the SCN^- in the mixture is as follows:

As we know that

$KSCN \rightarrow K^ + SCN^-$

As it is mentioned in the question that KSCN is present 10 mL of 0.05 M

So, the total milimoles of SCN^- is

= 10 × 0.05

= 0.5 m moles

The total volume in mixture is

= 45 + 10 + 30

= 85 mL

Now the initial concentration of the SCN^- is

= 0.5 ÷ 85

= 0.00588 M

hence, the initial concentration of SCN- in the mixture is 0.00588 M

5 0

3 years ago

The molarity of an aqueous sodium phosphate solution is 0.650 M. What is the molality of sodium ions present in this solution? T

netineya [11]

Answer:

molality of sodium ions is 1.473 m

Explanation:

Molarity is moles of solute per litre of solution

Molality is moles of solute per kg of solvent.

The volume of solution = 1 L

The mass of solution = volume X density = 1000mL X 1.43 = 1430 grams

The mass of solute = moles X molar mass of sodium phosphate = 0.65X164

mass of solute = 106.6 grams

the mass of solvent = 1430 - 106.6 = 1323.4 grams = 1.3234 Kg

the molality = $\frac{moles of solute}{mass of solvent in kg}=\frac{0.65}{1.323}= 0.491m$

Thus molality of sodium phosphate is 0.491 m

Each sodium phosphate of molecule will give three sodium ions.

Thus molality of sodium ions = 3 X 0.491 = 1.473 m

7 0

3 years ago