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2 years ago

How is density found

Chemistry

2 answers:

topjm [15]2 years ago

5 0

(m*v) is the result of it

Yanka [14]2 years ago

3 0

Through mass and volume.

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Make a poem about compounds

OleMash [197]

Answer:

<h3>A Werewolf</h3>

___________

Poem with Compound Words

------------------------------------------

whenever there's a full moon,

I cannot overlook,

some alterations in my ways,

and changes in my look.

my werewolf hair grows everywhere,

my werewolf teeth get long,

my eyesight gets much keener and

I'm muscular and strong.

I get to roam around outside,

the moonlight makes me howl,

these otherworldly sound effects

mean I am on the prawl.

I see the moon is round and full,

I'm moonstruck by the sight,

I've made some telltale changes-so, you'd best stay in tonight.

Explanation:

I HOPE IT HELPS YOUR ENGLISH SUBJECT ;) ★

4 0

2 years ago

What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11

lutik1710 [3]

The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m

4 0

3 years ago

What color are the stars on the diagram with the lowest surface temperatures?

Luba_88 [7]

Answer:

red supergiants is the answer

3 0

2 years ago

A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt

Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

$dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt$

integrating factor

$=e^{\int\limits (1/40) \, dt}$

integrating factor $=e^{(1/40)t}$

multiply on both sides by $=e^{(1/40)t}$

$dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t$

integrate on both sides

$\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})$

after long period of time means t - > ∞

${t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80$

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

6 0

3 years ago

Help question 9 is due today also :)

devlian [24]

Answer:

Car 3 with a net force of 12N

Explanation:

The formula is F=MA

Hope this helps friend

8 0

3 years ago