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jarptica [38.1K]
2 years ago
11

Write the three one of middle ear. class 9​

Chemistry
1 answer:
Paha777 [63]2 years ago
6 0

Answer:

The middle ear contains three tiny bones known as the ossicles: malleus, incus, and stapes. The ossicles were given their Latin names for their distinctive shapes; they are also referred to as the hammer, anvil, and stirrup, respectively.

Explanation:

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When might neutralization reactions be used in a laboratory setting?
Ipatiy [6.2K]
Neutralization reactions can be used in a laboratory setting in order t<span>o dispose of chemicals. When spills happens, for instance an acid is on the floor, you can use a base to neutralize the spill. Hope this answers the question. Have a nice day.</span>
6 0
2 years ago
How much thermal energy is added to 10.0 g of ice at −20.0°C to convert it to water vapor at 120.0°C?
Sonbull [250]

Answer:

7479 cal.

31262.2 joules

Explanation:

This is a calorimetry problem where water in its three states changes from ice to vapor.

We must use, the calorimetry formula and the formula for latent heat.

Q = m . C . ΔT

Q = Clat . m

First of all, let's determine the heat for ice, before it melts.

10 g . 0.5 cal/g°C ( 0° - (-20°C) = 100 cal

Now, the ice has melted.

Q = Clat heat of fusion . 10 g

Q = 79.7 cal/g . 10 g → 797 cal

We have water  at 0°, so this water has to receive heat until it becomes vapor. Let's determine that heat.

Q = m . C . ΔT

Q = 10 g . 1 cal/g°C (100°C - 0°C) → 1000 cal

Water is ready now, to become vapor so let's determine the heat.

Q = Clat heat of vaporization . m

Q = 539.4 cal/g . 10 g → 5394 cal

Finally we have vapor water, so let's determine the heat gained when this vapor changes the T° from 100°C to 120°

Q = m . C . ΔT

Q = 10 g . 0.470 cal/g°C . (120°C - 100°C) → 94 cal

Now, we have to sum all the heat that was added in all the process.

100 cal + 797 cal + 1000 cal + 5394 cal + 94 cal =7479 cal.

We can convert this unit to joules, which is more acceptable for energy terms.

1 cal is 4.18 Joules.

Then, 7479 cal are (7479 . 4.18) = 31262.2 joules

6 0
3 years ago
What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?
Sergeu [11.5K]

Answer:

\large \boxed{\text{b. 1.5 h}}

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463  h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}

4 0
3 years ago
How many moles of HCI would be required to produce a total of 2 moles of H2
k0ka [10]

Answer:

Should be 4 moles.

4 0
3 years ago
True or False? The following set of quantum numbers is acceptable (0, 0, 0, +1/2). Explain.
leonid [27]

Answer:

False

Explanation:

The values of the principal quantum number range from 1- infinity. This implies that there is no such thing as an energy level with n=0.

Hence this can not be an acceptable set of quantum numbers when n=0.

8 0
3 years ago
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