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3 years ago

Hydrogen cyanide (HCN) is a product of combustion that:

Physics

1 answer:

bekas [8.4K]3 years ago

5 0

b. is equally as toxic as carbon monoxide (CO).

Explanation:

Hydrogen Cyanide is a product of combustion that is equally as toxic as carbon monoxide CO.

The gas is a product of combustion in blast furnances, coke ovens, e.t.c

Just like carbon monoxide, they are blood agents that are toxic in the body
Prolonged exposure to hydrogen cyanide can lead to death eventually.
The gas is a able to produce cyanide ion that seriously impacts the process of cellular respiration and bringing it to a halt.
It is listed as one of the dangerous chemical weapons.

Learn more:

Safety data sheets brainly.com/question/2188622

#learnwithBrainly

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Light is reflected from a crystal of table salt with an index of refraction of 1.544. An analyser is placed to intercept the ref

Vitek1552 [10]

Answer:

hola me llamo bruno y tu?

Explanation:

pero yo soy de mexico

5 0

3 years ago

Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a specific enthalpy of 2431.6 kJ/kg and exi

PIT_PIT [208]

Answer:

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

hf = 376.57 kJ/kg

hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

= 0.025 × (2431.552 - 376.57)

= 0.025 × 2055.042

= 51.37455 kW

= 51.38 kW.

5 0

3 years ago

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm . The explorer finds that

larisa86 [58]

Answer:

12.4 m/s²

Explanation:

L = length of the simple pendulum = 53 cm = 0.53 m

n = Number of full swing cycles = 99.0

t = Total time taken = 128 s

T = Time period of the pendulum

g = magnitude of gravitational acceleration on the planet

Time period of the pendulum is given as

$T = \frac{t}{n}$

$T = \frac{128}{99}$

T = 1.3 sec

Time period of the pendulum is also given as

$T = 2\pi \sqrt{\frac{L}{g}}$

$1.3 = 2(3.14) \sqrt{\frac{0.53}{g}}$

g = 12.4 m/s²

4 0

3 years ago

Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has magnitude of 3.7 m/s^2. We want

denpristay [2]

Answer:

2.32 s

Explanation:

Using the equation of motion,

s = ut+g't²/2............................ Equation 1

Where s = distance, u = initial velocity, g' = acceleration due to gravity of the moon, t = time.

Note: Since Onur drops the basket ball from a height, u = 0 m/s

Then,

s = g't²/2

make t the subject of the equation,

t = √(2s/g')...................... Equation 2

Given: s = 10 m, g' = 3.7 m/s²

Substitute this value into equation 2

t = √(2×10/3.7)

t = √(20/3.7)

t = √(5.405)

t = 2.32 s.

4 0

3 years ago

Read 2 more answers

Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

AURORKA [14]

Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

3 0

3 years ago