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const2013 [10]
2 years ago
14

The liquid in your stomach has a pH of about 2, so it is ________.I

Chemistry
2 answers:
iren2701 [21]2 years ago
8 0

Answer:

A. an acid

Explanation:

Taya2010 [7]2 years ago
7 0

Answer:A because I said so.

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Someone help me with chemistry on my account/page please :))
Gemiola [76]

Answer:

ok

Explanation:

5 0
2 years ago
There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In
KIM [24]

Answer:

ΔH = -470.4kJ

Explanation:

It is possible to sum 2 or more reactions to obtain the ΔH of the reaction you want to study (Hess's law). Using the reactions:

1. CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)ΔH = −414kJ

2. 6C2H2(g) + 3CO2(g) + 4H2O(g) → 5CH2CHCO2H(g)ΔH = 132kJ

6 times the reaction 1.

6CaC2(s) + 12H2O(l) → 6C2H2(g) + 6Ca(OH)2(s)ΔH = −414kJ*6 = -2484kJ

This reaction + 2:

6CaC2(s) + 3CO2(g) + 16H2O(l) →  + 6Ca(OH)2(s) + 5CH2CHCO2H(g) ΔH = -2484kJ + 132kJ = -2352kJ

As we want to calculate the net change enthalpy in the formation of just 1 mole of acrylic acid we need to divide this last reaction in 5:

6/5CaC2(s) + 3/5CO2(g) + 16/5H2O(l) →  + 6/5Ca(OH)2(s) + CH2CHCO2H(g) ΔH = -2352kJ / 5

<h3>ΔH = -470.4kJ</h3>

4 0
3 years ago
If the conjugate base of a molecule has a pkb of 1.4, what would you expect the molecule to be?
maw [93]

If the conjugate base of a molecule has a pKb of 1.4, the molecule should be a Weak Acid.

Notice this question gives us the pKb of the molecule, not the pKa. Because of this, the pH scale basically gets reversed, so lower numbers in pKb correlate with stronger bases, and higher numbers in pKb correlate with stronger acids - the exact opposite of the pH scale.

It's important to make sure you completely understand the terms of conjugate base, conjugate acid, pKb, pKa, and how they all relate. It's easy to mix up the meanings of these definitions.

Here are the two other pieces of information you need to know to correctly answer this question:

  • Strong acids have a weak conjugate base.
  • Strong bases have a weak conjugate acid.

So if the problem says you have a strong conjugate base, then the molecule must be a weak acid. To illustrate this, think of ammonium, NH4+. Ammonium is a weak acid, but the conjugate base of ammonium is ammonia, NH3, which is a reasonably good base.

Learn more about conjugate base here : brainly.com/question/22514615

#SPJ4

3 0
2 years ago
What is the difference between hydrophilic and hydrophobic?
Nataliya [291]
Hydrophilic - attracted to water
Hydrophobic - resists water
7 0
2 years ago
Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.
Marianna [84]

The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
  • Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
  • Initial temperature (T₁) = –25 °C
  • Final temperature (T₂) = 0 °
  • Change in temperature (ΔT) = 0 – (–38) = 38 °C
  • Specific heat capacity (C) = 2050 J/(kg·°C)
  • Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
  • Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
  • Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 100 °C
  • Final temperature (T₂) = 160 °C
  • Change in temperature (ΔT) = 160 – 100 = 60 °C
  • Specific heat capacity (C) = 1996 J/(kg·°C)
  • Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
  • Heat for –38 °C to 0°C (Q₁) = 31160 J
  • Heat for melting (Q₂) = 133600 J
  • Heat for 0 °C to 100 °C (Q₃) = 167200 J
  • Heat for vaporization (Q₄) = 904000 J
  • Heat for 100 °C to 160 °C (Q₅) = 47904 J
  • Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

brainly.com/question/10286596

#SPJ1

7 0
2 years ago
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