A molecure that contains 3 iddentical polar bonds to the central atom will be

A scientist claims to have a cooling apparatus kept at -100C by liquid nitrogen. Is this possible? Why or why not?

Vilka [71]

Answer:

It is not possible

Explanation:

Nitrogen is a light gas that find application in my areas including being used to keep things cold.It has a freezing point of -210°c so the scientist Nitrogen in the question at 100°c have already freezed and no longer a liquid.

5 0

4 years ago

How many moles of glucose (C6H12O6) are in 4.0 liters of a 4.5 M C6H12O6 solution? (3 points) Question 1 options: 1) 18 moles 2)

qwelly [4]

You need to start by subbing in the correct numbers for these values/properties in this formula:

number of moles (mol) = molar concentration (mol/L) x volume (L)

Hope this helps! Have a great day!

6 0

4 years ago

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A solution is prepared by dissolving 0.26 mol of hydrazoic acid and 0.26 mol of sodium azide in water sufficient to yield 1.00 L

Pavel [41]

Answer:

The pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

Explanation:

The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:

HN₃(aq) + H₂O(l) ⇄ N₃⁻(aq) + H₃O⁺(aq)

The pH of this solution is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60$

When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:

H₃O⁺(aq) + N₃⁻(aq) ⇄ HN₃(aq) + H₂O(l)

The number of moles of NaN₃ after the reaction with HCl is:

$\eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles$

Now, the number of moles of HN₃ is:

$\eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles$

Then, the pH of the buffer solution after the addition of HCl is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43$

The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.

Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

I hope it helps you!

6 0

3 years ago

Why does the dilution equation, M 1 V 1 = M 2 V 2, work?

valentina_108 [34]

Answer:

d) The dilution equation works because the number of moles remains the same.

Explanation:

Let’s say that you have 1 mol of a solute in I L of solution. The concentration is 1 mol·L⁻¹. and M₁V₁ = 1 mol.

Now, you dilute the solution to a volume of 2 L. You still have 1 mol of solute, but in 2 L of solution. The new concentration is 0.5 mol·L⁻¹.

The volume has doubled, but the volume has halved, and M₂V₂ = 1 mol.

b) Wrong. The molar concentration changes on dilution.

c) Wrong. The volume changes on dilution.

a) Wrong, although technically correct, because if the moles don’t change, the mass doesn’t change either. However, the formula M₁V₁ has units mol·L⁻¹ × L = mol. Thus, in the formula, it is moles that are constant.

6 0

4 years ago

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Describe two uses of surfactants. At least one must be something that was not described in the passage. For each example, explai

notka56 [123]

There are several types of surfactants, to name two: alkylbenzene sulfonates or detergents; and lauryl sulfate or foaming agents.

Alkylbenzene sulfonates are commonly used in the production of detergent soaps, these help clean the dirt from cloths and other materials, it also produces foam.

Lauryl sulfate or foaming agents are commonly used in the production of dishwashing liquid or handwashing soap.

8 0

4 years ago

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