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4 years ago

What is the missing part of the chemical equation shown below? acid + base ? salt + ________

Chemistry

2 answers:

Gennadij [26K]4 years ago

5 0

Water is the answer.

lions [1.4K]4 years ago

3 0

Water
example HCl + NaOH= NaCl +H2O

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According to the Arrhenius equation, changing which factors will affect the

slamgirl [31]

Answer:

e−(Ea/RT): the fraction of the molecules present in a gas which have energies equal to or in excess of activation energy at a particular temperature

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3 years ago

The brightness of a star is determined by _____.

Travka [436]

Answer:

Size and Temperature or E & B

Explanation:

7 0

2 years ago

Read 2 more answers

For each of the following compounds, identify what type of bonding holds them together.

svet-max [94.6K]

Explanation

NaCl: Ionic crystal lattice forces

Hg: Metallic bonding

CO₂: London dispersion forces

CH₄: London dispersion forces

Li₂O: Ionic crystal lattice forces

Ag: Metallic bonds

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4 0

3 years ago

0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.o

Lady bird [3.3K]

Answer:

10.57% of N in acetanilide

Explanation:

All nitrogen in the sample is converted in NH₃ in the Kjeldahl determination. The NH₃ reacts with H₂SO₄ as follows:

2NH₃ + H₂SO₄ → 2NH₄⁺ + SO₄²⁻

The acid in excess in titrated with Na₂CO₃ as follows:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

To solve this question we must find the moles of sodium carbonate = Moles of H₂SO₄ in excess. The added moles - Moles in excess = Moles of sulfuric acid that reacts:

Moles Na₂CO₃ anf Moles H₂SO₄ in excess:

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

Moles H₂SO₄ added:

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

Moles that react:

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

Moles of NH₃ = Moles N:

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

mass N and mass percent:

0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =

<h3>10.57% of N in acetanilide</h3>

7 0

3 years ago

Salt in crude oil must be removed before the oil undergoes processing in a refinery. The

irina1246 [14]

Answer:

$\large \boxed{0.64 \, \%}$

Explanation:

Assume you are using 1 L of water.

Then you are washing 4 L of salty oil.

1. Calculate the mass of the salty oil

Assume the oil has a density of 0.86 g/mL.

$\text{Mass of oil} = \text{4000 mL} \times \dfrac{\text{0.86 g}}{\text{1 mL}} = \text{3440 g}$

2. Calculate the mass of salt in the salty oil

$\text{Mass of salt} = \text{3440 g} \times \dfrac{\text{5 g salt}}{\text{100 g oil}} = \text{172 g salt}$

3. Calculate the mass of salt in the spent water

$\text{Mass of salt} = \text{1000 g water} \times \dfrac{\text{15 g salt}}{\text{100 g water}} = \text{150 g salt}$

4. Mass of salt remaining in washed oil

Mass = 172 g - 150 g = 22 g

5. Concentration of salt in washed oil

$\text{Concentration} = \dfrac{\text{22 g}}{\text{3440 g}} \times 100 \, \% = \mathbf{0.64 \, \%}\\\\\text{The concentration of salt in the washed oil is $\large \boxed{\mathbf{0.64 \, \%}}$}$

3 0

3 years ago