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3 years ago

Can you substitute evaporated milk for condensed milk

1 answer:

3 0

Yes because condensed milk and evaporated milk are similar to one another. However, there won’t be the same sweet flavor but the texture is the same.

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example best shows that the chemistry of water is helpful to plants? Water’s polarity produces a high density, which allows wate

Gennadij [26K]

The right answer for the question that is being asked and shown above is that: "Water’s polarity produces a high density, which allows water to move to the leaves." Example best shows that the chemistry of water is helpful to plants is that <span>Water’s polarity produces a high density, which allows water to move to the leaves.</span>

8 0

3 years ago

Read 2 more answers

Equal masses (in grams) of hydrogen gas and oxygen gas are reacted to form water. Which substance is limiting?

Sonbull [250]

Answer:

a. Oxygen gas is limiting

Explanation:

hydrogen gas and oxygen gas are reacted to form water

2H₂ + O₂ → 2H₂O

the above balanced equation shows that 2 moles of H₂ is required for 1 mole of O₂

Given equal masses of H₂ and O₂

assuming 'x' gm for each, no. of moles of each gas =

no. of moles of H₂ = x/2 = 0.5x moles

no.of moles of O₂ = x/32 = 0.031x moles

This shows that no. of moles of O₂ is very less so O₂ will become the limiting reagent.

3 0

3 years ago

Read 2 more answers

Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid

Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

$35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol$

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

$\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M$

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb) <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

7 0

3 years ago

What is the volume of 1.56 kg of a compound whose molar mass is 81.86 g/mole and whose density is 41.2 g/ml?

hjlf

Answer:

v = 37.9 ml

Explanation:

Given data:

Mass of compound = 1.56 kg

Density = 41.2 g/ml

Volume of compound = ?

Solution:

First of all we will convert the mass into g.

1.56 ×1000 = 1560 g

Formula:

D=m/v

D= density

m=mass

V=volume

v = m/d

v = 1560 g / 41.2 g/ml

v = 37.9 ml

7 0

3 years ago

Find the partial pressure of helium gas in mmHg

mezya [45]

Answer:

214 mmhg

Explanation:

5 0

2 years ago