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2 years ago

Thin, flattened, and typically curved bones, such as the ribs and sternum, form __________ bones.

2 answers:

4 0

Thin, flattened, and typically curved bones, such as the ribs or sternum, form falt bones.

The two main functions of flat bones are protection or provide a large surface for musular attachment.

These are the flat bones of the human body: in the skull, parietal, frontal, occipial, lacrimal, nasal and vomer; scapula; sternum; rib and hip.

Snowcat [4.5K]2 years ago

4 0

Flat bones is the answer

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4) What volume will the gas in the balloon at right occupy at 250k? balloon: 4.3L 350K

swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

$Pv =nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 4.3 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

4.3 liter = $4.3\times10^{-3}=4.3\times10^{-3} m^{3}$

And initial temperature of balloon, $T_{1}$ = 350 K

Let the final volume of balloon is $V_{2}$

And as given, final temperature of balloon, $T_{2}$ is 250 K

Now, $V_{1} = KT_{1}$

$4.3\times10^{-3}=K\times350\ (equation\ 1 )$

$V_{2} = KT_{2}$

$=K\times250\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}$

K cancelled by K.

By cross multiplication:

$350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\ = \frac{1075\times10^{-3}}{350} \\ =2.87\times10^{-3}m^{3}$

Now, convert it into liter with the help of calculation done above,

$2.87\times10^{-3} \times1000\\2.87\times10^{-3} \times10^{3} \\2.87\ liter$

Therefore, volume of the gas in the balloon at 250 K will be 2.87 liter.

4 0

3 years ago

1. The most basic units of any element are its

baherus [9]

Answer:

The atom

Explanation:

5 0

2 years ago

A child has a toy balloon with a volume of 1.80 L. The temperature of the balloon when it was filled was 293 K at a pressure of

Mekhanik [1.2K]

Answer:

2.42L

Explanation:

Given parameters:

V₁ = 1.8L

T₁ = 293K

P₁ = 101.3kPa

P₂ = 67.6kPa

T₂ = 263K

Unknown:

V₂ = ?

Solution:

To solve this problem, we are going to use the combined gas law to find the final volume of the gas. The combined gas law expression combines the equation of Boyle's law, Charles's law and Avogadro's law;

$\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

All the units are in the appropriate form. We just substitute and solve for the unknown;

101.3 x 1.8 / 293 = 67.6 x V₂ / 263

V₂ = 2.42L

7 0

2 years ago

The energy source used in nuclear power plants today is

Klio2033 [76]

Answer:

B I have taken the quiz already 90%

7 0

3 years ago

This graph shows two curves pertaining to a hydrogen s orbital.

fgiga [73]

Answer 1) : According to the complete question attached in the answer,

The radial wave function which is denoted by $R_{nl}(r)$ shown with orange color crosses through zero point. Also, At the the radial nodes, which are spherical shells to some radial distance away from the nucleus there no electron are found.

Also, the radial probability distribution curve denoted as $R^{2}_{nl}(r)$ shown in blue color is observed to touch zero, and shows the place of radial node.

Therefore, the total number of nodes will include both the kinds which has radial and angular nodes which will be represented by 'n'.

It is observed that for any atomic orbital, the total number of nodes will be n-1 .

Considering the s orbital of the hydrogen, which has zero angular momentum (l); (l=0), as it has zero angular nodes.

Hence, there will be only radial nodes, which is

(n−1 = total number of radial nodes in s orbitals)

According to the image, there are 4 radial nodes shown, so n = 5 (as n-1 = 4; therefore, n = 5)

This represents the 5s orbital.

Answer 2) The radial nodes are observed in I'm seeing radial nodes at

$1.9a_{0}$ , $6.4a_{0}$ , $13.9a_{0}$ and $27.0a_{0}$ .

where $a_{0}$ represents the hydorgen bohr atomic radius = 0.0529177 nm

Explanation : It is quite easy to observe the given graph and find out the approximate values of the radial nodes, it does not requires any equation to be solved. Equation can be used to find the radial nodes if it was supplied along with the question. Although by mere speculation one can find out the answer.

3 0

2 years ago

Read 2 more answers