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3 years ago

A neutralization reaction between an acid and a metal hydroxide produces ________.

2 answers:

5 0

A neutralization reaction between an acid and a metal hydroxide will produce water and a salt. Consider the reaction between sodium hydroxide and hydrochloric acid. This would be:
NaOH + HCl --> NaCl + H2O
The resulting products are table salt (NaCl) and water (H2O)

irina [24]3 years ago

5 0

Answer : The correct option is, (a) water and a salt

Explanation :

Neutralization reaction : It is a type of chemical reaction in which an acid react with a base or metal hydroxide to give salt and water as a product that means it reacts to give a neutral solution.

For example :

When sodium hydroxide base react with hydrochloric acid then it react to gives sodium chloride and water as a product.

The balanced chemical reaction will be:

$HCl+NaOH\rightarrow NaCl+H_2O$

Hence, the correct option is, (a) water and a salt

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An oxidation reaction occurs only in the presence of _____.

aev [14]

The answer is reduction reaction. For every oxidation reaction that takes place, a reduction reaction must also happen, no reduction can occur without oxidation and vice versa. In redox reaction, atoms or ions undergo changes in the electronic structure, involving the transfer of electrons between reactants.

5 0

3 years ago

How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so

Svetach [21]

Answer:

V H2O = 170.270 mL

Explanation:

QH2O ( heat gained) = Qcoffe ( heat ceded)

⇒ Q = mCΔT

∴ m: mass (g)

∴ C: specific heat

assuming:

δ H2O = δ Coffe = 1.00 g/mL
C H2O = C coffe = 4.186 J/°C.g....from literature

⇒ Q coffe = (mcoffe)(C coffe)(60 - 95)

∴ m coffe = (180mL)(1.00 g/mL) = 180 g coffe

⇒ Q = (180g)(4.186 J/°C.g)(-35°C) = - 26371.8 J

⇒ Q H2O = 26371.8 J = (m)(4.186 J/°C.g)(60 - 23)

⇒ (26371.8 J)/(154.882 J/g) = m H2O

⇒ m H2O = 170.270 g

⇒ V H2O = (170.270 g)(mL/1.00g) = 170.270 mL

5 0

3 years ago

Why are changes in phase physical changes?

ehidna [41]

Weathering and chemical substance weathering

3 0

3 years ago

23.495 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 72.311 g of wate

Otrada [13]

Answer:

1.86% NH₃

Explanation:

The reaction that takes place is:

HCl(aq) + NH₃(aq) → NH₄Cl(aq)

We calculate the moles of HCl that reacted, using the volume used and the concentration:

32.27 mL ⇒ 32.27/1000 = 0.03227 L

0.1080 M * 0.03227 L = 3.4852x10⁻³ mol HCl

The moles of HCl are equal to the moles of NH₃, so now we calculate the mass of NH₃ that was titrated, using its molecular weight:

3.4852x10⁻³ mol NH₃ * 17 g/mol = 0.0592 g NH₃

The weight percent NH₃ in the aliquot (and thus in the diluted sample) is:

0.0592 / 12.949 * 100% = 0.4575%

Now we calculate the total mass of NH₃ in the diluted sample:

Diluted sample total mass = Aqueous waste Mass + Water mass = 23.495 + 72.311 = 95.806 g

0.4575% * 95.806 g = 0.4383 g NH₃

Finally we calculate the weight percent NH₃ in the original sample of aqueous waste:

0.4383 g NH₃ / 23.495 g * 100% = 1.86% NH₃

6 0

3 years ago

a helium balloon has a volume of 2.95 liters at 25 c. the volume of the ballon decreased to 2.25 l after it is placed outside on

OleMash [197]

Answer:

The outside temperature is -45.8°C

Explanation:

When a gas keeps on constant its moles and its pressure, we can assume that volume will be increased or decreased as the T° (absolute T° in K).

V1 / T1 = V2 / T2

2.95L/298K = 2.25L / T2

(2.95L/298K ) . T2 = 2.25L

T2 = 2.25L . 298K / 2.95L

T2 = 227.2K

T°K - 273 = T°C

227.2K - 273 = -45.8°C

3 0

4 years ago