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3 years ago

what gas was given off when hydrogen peroxide is broken down? recall that the formula for hydrogen peroxide is h2o2?

Chemistry

1 answer:

Anna35 [415]3 years ago

5 0

When hydrogen peroxide is broken down the gas released or given off is Oxygen.

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How do electrons in a metallic bond behave?

igomit [66]

Hello,

In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “sea” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.

Hope this helps!!!! Happy Holidays!!!! (:

4 0

3 years ago

Calculate weight of anhydrous compound from remains upon dehydration of 2g copper sulphate pentahydrate

Greeley [361]

Answer:

1.28 g

Explanation:

Mass of anhydrous compound/molar mass of anhydrous compound = mass of hydrated compound/ molar mass of hydrated compound

Mass of anhydrous compound = ?

Mass of hydrated compound = 2g

Molar mass of anhydrous compound= 160 g/mol

Molar mass of hydrated compound = 250 g/mol

x/160 = 2/250

250x = 2 ×160

x= 2 × 160/250

x= 1.28 g

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2 years ago

1 punto Kelvin, es la unidad de medida de: Intensidad o luminosa O Temperatura termodinámica O Corriente eléctrica O Cantidad de

Len [333]

Answer:

sorry i donntttt understand make it clear

7 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

2 years ago

What would you ask the scientists that came up with the big bang theory?

faust18 [17]

Answer:

What is the name of the drug you were talking at the time of coming up with this theory please?

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2 years ago