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umka21 [38]
3 years ago
8

If a patient\'s blood pressure is 145 over 65 mmHg, what is it in atmospheres (atm)

Chemistry
2 answers:
kicyunya [14]3 years ago
7 0

Converting mmHg to atm is solved by division.

Example: Convert 745.0 to atm.

Solution- divide the mmHg value by the 760.0 mmHg / atm.

745 mmHg over 760.0 mmHg/atm

atm value is 0.980263

Now, I am a medical student and we have never had to convert a BP (blood pressure) to atm from mmHg, only ever kPA. SO, I am going to take a guess here and say that when you do the work to solve this, you are going to convert the Systolic (upper #) which is the 145. You should get 0.190789 and then convert the Diastolic (lower #) which is 65. You should get 0.08552632.

So your fraction so to speak should read, 0.190789/0.08552632 or 0.190789 over 0.08552632

(Just to note that is way to low of a BP, although it is irrelevant) Best wishes and good luck. "Remember, never just look for the right answer, look for why it is the right answer!"

Andreyy893 years ago
6 0

Answer:

0.191/0.086 atm

Explanation:

Blood pressure is expressed with two numbers:

The first number is the systolic pressure which is the pressure when the heart beats.

The second number is the diastolic pressure which is the pressure when the heart rests between beats.

So, we must convert the two numbers to atm. Since 1 atm is equal to 760 mmHg we have:

145 mmHg x (1 atm/760 mmHg) = 0.191 atm

65 mmHg x (1 atm/760 mmHg) = 0.086 atm

The patient’s blood pressure is 0.191/0.086 atm

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1 year ago
slader Consider the following reactions: A. uranium-238 emits an alpha particle; B. plutonium-239 emits an alpha parti- cle; C.
pishuonlain [190]

<u>Answer:</u>

<u>For A:</u> The equation is _{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha

<u>For B:</u> The equation is _{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha

<u>For C:</u> The equation is _{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta

<u>Explanation:</u>

Alpha decay process is the process in which nucleus of an atom disintegrates into two particles. The first one which is the alpha particle consists of two protons and two neutrons. This is also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.

_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha

Beta decay process is defined as the process the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

<u>For A:</u> Uranium-238 emits an alpha particle

The nuclear equation for this process follows:

_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha

<u>For B:</u> Plutonium-239 emits an alpha particle

The nuclear equation for this process follows:

_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha

<u>For C:</u> Thorium-239 emits a beta particle

The nuclear equation for this process follows:

_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta

6 0
3 years ago
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The electrochemical equivalent of copper, Cu, is 3.29015544 × 10⁻⁷ g/C

Explanation:

The given parameters are;

The element for which the electrochemical equivalent is sought = Copper

The atomic mass of copper = 63.5

The electrochemical equivalent, 'Z', of an element or a substance is the mass, 'm', of the element or substance deposited by one coulomb of electricity, which is equivalent to a 1 ampere current flowing for a period of 1 second

Mathematically, we have;

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We have;

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Therefore, one mole of Cu, is deposited by 2 moles of electrons

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∴ The charge carried two moles of electrons, Q = 2 × 96500 C = 193,000 C

Given that the mass of an atom of Cu = 63.5 a.m.u., the mass of one mole of Cu, m = 63.5 g

Z = \dfrac{m}{Q} = \dfrac{63.5 \ g}{193,000 \ C} = 3.29015544 \times 10^{-4} \, g \cdot C^{-1}

∴ Z = 3.29015544 × 10⁻⁴ g/C = 3.29015544 × 10⁻⁷ g/C

The electrochemical equivalent of copper, Cu, is Z = 3.29015544 × 10⁻⁷ g/C

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