3 years ago

From which celestial body is a star born?

Chemistry

2 answers:

shepuryov [24]3 years ago

5 0

Answer:

B. nebula

Explanation:

Birth takes place inside <u>hydrogen-based</u> dust clouds called nebulae.

<u>-TheUnknownScientist</u>

dolphi86 [110]3 years ago

4 0

Answer: B. nebula

Explanation:

You might be interested in

if the half-life of substance A is 1.5 years, how long would it take a 8.0 gram sample of substance A to decay such that only 1.

Fudgin [204]

Answer:

4.5

Explanation:

3 0

3 years ago

Rock debris from space called enter Earth and are called . If they survive this fall and land on Earth, they are called .

Serjik [45]

Meteorites, I think it’s an asteroid when in space and a meteorite once it lands

8 0

3 years ago

Read 2 more answers

Water supplies are often treated with chlorine as one of the processing steps in treating wastewater. Estimate the liquid diffus

jekas [21]

Answer:

⇒D_AB= 1.21×10^(-9)

Explanation:

Wike chang equation is given as:

$D_{AB}= \frac{117.3\times10^{-18}\times\(\phi\times M_B)^{0.5}\times T}{\mu\times\nu^{0.6}}$

Where

D_AB= diffusivity of chlorine in water

Φ= 2.26 for water as solvent

ν= 0.0484 for chlorine as solute

M_B = Molecular weight of water

τ= temperature=289 K

μ= viscosity = 1.1×10^{-3}

Now putting these values in the above equation we get

$D_{AB}= \frac{117.3\times10^{-18}\times\(\2.26\times18)^{0.5}\times289}{\1.1\times10^{-3}\times\0.0484^{0.6}}$

⇒D_AB= 1.21×10^(-9)

7 0

3 years ago

Read 2 more answers

Mercury, which is sometimes used in thermometers, has a density of 13.534 g/ml at room temperature. what volume of mercury conta

Sergeu [11.5K]

D - density: 13,534 g/ml
m - mass: 10g
V - volume: ??
_____________
d = m/V
V = m/d
V = 10/13,534
V = 0,7389 ml

:•)

6 0

4 years ago

If the rate of decrease for the partial pressure of N2H4N2H4 in a closed reaction vessel is 70 torr/htorr/h , what is the rate o

german

Answer:

$r_{NH_3}=140torr/h$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$N_2H_4 (g) + H_2 (g) \rightarrow 2 NH_3$

Thus, in terms of pressures, the rate becomes:

$-r_{N_2H_4}=\frac{1}{2} r_{NH_3}$

Thus, the rate of change for the partial pressure of ammonia turns out:

$r_{NH_3}=2*(-r_{N_2H_4})\\r_{NH_3}=2*[-(-70torr/h)]\\r_{NH_3}=140torr/h$

The rate of decrease of partial pressure of urea is taken negative as it is a reactant whereas ammonia a product which has 2 as its stoichiometric coefficient.

Best regards.

7 0

3 years ago