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Kobotan [32]
3 years ago
13

How many moles are 3.21×10^28 molecules of N2?

Chemistry
1 answer:
kow [346]3 years ago
8 0

Answer:5 to the 7th power

Explanation:

Divided 7 and y then add 4

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Based on the balanced chemical equation for the preparation of malachite, what is the composition of the bubbles formed when sod
spin [16.1K]

Answer:

Carbon Dioxide = CO2

Explanation:

The synthesis of Malachite is seen in the chemical formula:

CuSO 4 . 5H2O(aq) + 2NaCO3(aq) --> CuCO 3 Cu(OH) 2 (s) + 2Na 2 SO 4 (aq) + CO 2 (g) + 9H 2 O(l)

The bubbles mentioned in the question hints that our interest is the compounds in their gseous phase  (g).

Upon examining the chemical equation, only CO2 is in the gaseous state and hence the only one that can be formed as bubbles,

3 0
3 years ago
If i have 17 moles of gas at a temperature of 67c, and a volume of 88.89 liters what is the pressure of the gas
Shtirlitz [24]
Use the state equation for ideal gases: pV = nRT

Data:

V = 88.89 liter
n = 17 mol
T = 67 + 273.15 = 340.15 K

R = 0.0821 atm * liter / (K*mol)

=> p = nRT / V = 17 mol * 0.0821 (atm*liter / K*mol) * 340.15 K / 88.89 liter

p = 5.34 atm

Answer: p = 5.34 atm

4 0
3 years ago
Enough of a monoprotic weak acid is dissolved in water to produce a 0.0172 M solution. If the pH of the resulting solution is 2.
Mumz [18]

Answer:

9.7 x 10⁻⁴

Explanation:

              HA     ⇄           H⁺     +        A⁻

C(eq)   0.0174             10⁻²·³⁹         10⁻²·³⁹

                              =0.0041M     =0.0041M

Ka = [H⁺][A⁻]/[HA] = (0.0014)²/(0.0174) = 9.7 x 10⁻⁴

8 0
3 years ago
Which part of dalton theory did thomson finding dispute
rjkz [21]
Dalton thought that atoms were indivisible particles. But Thomsons discovery of the electron proved that the subatomic particles exist.
4 0
4 years ago
Tungsten (W) and chlorine (Cl) form a series of compounds with the following compositions:_______.
deff fn [24]

Answer:

WCl₂, WCl₄, WCl₅, WCl₆

Explanation:

Molar Mass of Tungsten = 184 g/mol

Mass of Chlorine = 35.5 g/mol

In the first compound;

Percentage of tungsten = 72.17 %

Upon solving;

72.17 % = 184

100 % = Total mass

Total mass of compound = 254.95g

Mass of chlorine = 254.95 - 184 = 70.95 (Dividing by 35.35; This is approximately 2 Chlorine atoms.

The Formular is WCl₂

In the second compound;

Percentage of tungsten = 56.45 %

Upon solving;

56.45 % = 184

100 % = Total mass

Total mass of compound = 325.95 g

Mass of chlorine = 325.95 - 184 = 141.95g (Dividing by 35.35; This is approximately 4 Chlorine atoms.

The Formular is WCl₄

In the third compound;

Percentage of tungsten = 50.91 %

Upon solving;

50.91 % = 184

100 % = Total mass

Total mass of compound = 361.42 g

Mass of chlorine =  361.42 - 184 = 177.42 (Dividing by 35.35; This is approximately 5 Chlorine atoms.

The Formular is WCl₅

In the fourth compound;

Percentage of tungsten = 46.39 %

Upon solving;

46.39 % = 184

100 % = Total mass

Total mass of compound = 396.64 g

Mass of chlorine = 396.64 - 184 = 212.64 (Dividing by 35.35; This is approximately 6 Chlorine atoms.

The Formular is WCl₆

4 0
3 years ago
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