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2 years ago

What would be the best flow rate of o2 for a patient with copd?

1 answer:

4 0

When we say COPD, this means Chronic Obstructive Pulmonary Disorder. COPD includes a group of lung diseases which have progressed into a worse condition. One example of this is emphysema. In cases of COPD, they should be provided oxygen therapy and the best flow rate for patients with this disorder is <span>8-10 L/min, high-flow, via nasal prong. Hope this helps.</span>

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If a helium filled ballon has a volume of 3.4l at 25°c and 1.18atm., what is its volume at STP?

vladimir2022 [97]

Answer:

3.711 L

Explanation:

The formula you need to use is the following:

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

3.4L / 298 K = V2 / 273 K

V2 = 3.711 L

5 0

2 years ago

The electrons formed from the aerobic oxidation of glucose are

SVEN [57.7K]

Answer:

i and ii

Explanation:

In the aerobic oxidation of glucose, the electrons formed are transferred to O2 after several others transfer reactions like passing through coenzymes NAD+ and FAD

4 0

2 years ago

Misconceptions about psychological disorders

exis [7]

Answer:

Explanation:

Misconception #1: Mental illnesses aren't real.

Misconception #2: Youth can't experience mental illness.

Misconception #3: Mental illness is a sign of weakness.

Misconception #4: People living with a mental illness will never get better

4 0

3 years ago

What distinguishes compounds from mixtures?

dexar [7]

Answer:

There's a very small difference between the two...a compound is when you mix several elements to create a whole new one. A mixture is when you mix several elements together, but instead of it creating a completely different compound, it's simply just a few elements mixed together.

3 0

3 years ago

You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of

Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$

Amount of BaSO₄ from K₂SO₄ is;

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$ = $X\times\frac{233.38 }{142.04}$ = 1.643·X

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$ = $Y\times\frac{233.38 }{174.259 }$ = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

8 0

3 years ago