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A 0.9% solution of NaCl is considered isotonic to mammalian cells. what molar concentration is this?

natka813 [3]

Answer:

58.44 g/mol The Molarity of this concentration is 0.154 molar

Explanation:

the molar mass of NaCl is 58.44 g/mol,

0.9 % is the same thing as 0.9g of NaCl , so this means that 100 ml's of physiological saline contains 0.9 g of NaCl. One liter of physiological saline must contain 9 g of NaCl. We can determine the molarity of a physiological saline solution by dividing 9 g by 58 g... since we have 9 g of NaCl in a liter of physiological saline, but we have 58 grams of NaCl in a mole of NaCl. When we divide 9 g by 58 g, we find that physiological saline contains 0.154 moles of NaCl per liter. That means that physiological saline (0.9% NaCl) has a molarity of 0.154 molar. We can either express this as 0.154 M or 154 millimolar (154 mM).

3 0

4 years ago

20. Antifreeze spills should be contained with:

denis23 [38]

Answer:

D

Explanation:

the sponge's and rags are meant for this exact purpose

3 0

3 years ago

For each topic, find the total number of blurts that were analyzed as being related to the topic. Order the result by topic id.

photoshop1234 [79]

Answer:

Explanation: see attachment below

8 0

3 years ago

Determine the resolution of a manometer required to measure the velocity of air at 50 m/s using a pitot-static tube and a manome

oksano4ka [1.4K]

Answer:

a) Δh = 2 cm, b) Δh = 0.4 cm

Explanation:

Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used in such a way that the height of point 2 of is the same of point 1

y₁ = y₂

subtitute

P₁ = P₂ + ½ ρ v₂²

P₁ -P₂ = ½ ρ v²

where ρ is the density of fluid

now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface

ΔP =ρ_{Hg} g h - ρ g h

ΔP = (ρ_{Hg} - ρ) g h

as the static pressure we can equalize the equations

ΔP = P₁ - P₂

(ρ_{Hg} - ρ) g h = ½ ρ v²

v = $\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}$

in this expression the densities are constant

v = A √h

A = $\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }$

They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³

we look for the constant

A = $\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }$

A = 454.55

we substitute

v = 454.55 √h

to calculate the uncertainty or error of the velocity

h = $\frac{1}{454.55^2} \ v^2$

Δh = $\frac{dh}{dv}$ Δv

$\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}$

Suppose we have a height reading of h = 20 cm = 0.20 m

a) uncertainty 2.5 m / s ( 0.05)

$\frac{\delta v}{v} = 0.05$

$\frac{\Delta h}{h}$ = 2 0.05

Δh = 0.1 h

Δh = 0.1 20 cm

Δh = 2 cm

b) uncertainty 0.5 m / s ( Δv/v= 0.01)

$\frac{\Delta h}{h}$ = 2 0.01

Δh = 0.02 h

Δh = 0.02 20

Δh = 0.1 20 cm

Δh = 0.4 cm = 4 mm

5 0

3 years ago

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude

Fofino [41]

Answer:

$u = 260.22m/s$

$S_{max} = 1141.07ft$

Explanation:

Given

$S_0 = 89.6ft$ --- Initial altitude

$S_{16.5} = 0ft$ -- Altitude after 16.5 seconds

$a = -g = -32.2ft/s^2$ --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

$S = ut + \frac{1}{2}at^2$

The final altitude after 16.5 seconds is represented as:

$S_{16.5} = S_0 + ut + \frac{1}{2}at^2$

Substitute the following values:

$S_0 = 89.6ft$ $S_{16.5} = 0ft$ $a = -g = -32.2ft/s^2$ and $t = 16.5$

So, we have:

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2$

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45$

$0 = 89.6 + 16.5u- 4383.225$

Collect Like Terms

$16.5u = -89.6 +4383.225$

$16.5u = 4293.625$

Make u the subject

$u = \frac{4293.625}{16.5}$

$u = 260.21969697$

$u = 260.22m/s$

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

$v=u + at$

At the maximum height:

$v =0$ --- The final velocity

$u = 260.22m/s$

$a = -g = -32.2ft/s^2$

So, we have:

$0 = 260.22 - 32.2t$

Collect Like Terms

$32.2t = 260.22$

Make t the subject

$t = \frac{260.22}{ 32.2}$

$t = 8.08s$

The maximum height is then calculated as:

$S_{max} = S_0 + ut + \frac{1}{2}at^2$

This gives:

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2$

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22$

$S_{max} = 89.6 + 260.22 * 8.08 - 1051.11$

$S_{max} = 1141.0676$

$S_{max} = 1141.07ft$

Hence, the maximum height is 1141.07ft

8 0

3 years ago