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3 years ago

How many mole are in 2.84*10^22 molecules of h2

1 answer:

4 0

Avogadro's Number and the Mole. The mole is represented by Avogadro's number, which is 6.022×1023 atoms or molecules per mol.

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A lahar is a type of mudflow that occurs

Fittoniya [83]

I'd guess A or D, but my gut says A

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3 years ago

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Which elements have the most similar chemical properties?

bixtya [17]

Answer:

a. Be and Mg

Explanation:

Be (beryllium) and Mg (magnesium) are both found in the alkaline earth metals family (group 2) on the periodic table. Elements in the same family or group have similar properties,

Ca and Br are in groups 2 and 17

Cl and Ar are in groups 17 and 18

Na and P are in groups 1 and 15

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3 years ago

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Consider a pot of water at 100 C. If it took 1,048,815 J of energy to vaporize the water and heat it to 135 C, how many grams of

jeka57 [31]

Answer:

There was 450.068g of water in the pot.

Explanation:

Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L

Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s

Let m = x g be the weight of water in the pot.

Energy required to vaporise water = mL = 2260x

Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x

Total energy required = $2260x+x\times2.010\times(135-100)=2260x+70.35x=2330.35x$

$2330.35x=1048815\\x=450.068g$

Hence, there was 450.068g of water in the pot.

8 0

2 years ago

Which of the following is the SI unit used to measure mass? a. kilogram b. liter c. meter d. Kelvin Please select the best answe

Ket [755]

Answer: d

Explanation: because it’s right

5 0

3 years ago

When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot

viktelen [127]

Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the $'n^{th}'$ orbit of hydrogen atom.

Formula used :

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV$

Energy of n = 2 in an hydrogen atom:

$E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV$

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

$E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV$

The negative sign indicates that energy of the photon emitted.

Thus, the energy of the photon emitted is, -12.1 eV

3 0

2 years ago