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kiruha [24]
1 year ago
6

How many moles of H2O are in 64.0 g of H2O

Chemistry
1 answer:
Alina [70]1 year ago
8 0

Answer:

Explanation:

stoichiometry of C₂H₂ to H₂O is 2:2.

Number of moles of C₂H₂ = molar mass of C₂H₂  

Since the molar mass of C₂H₂  is 26 g/mol.

Number of C₂H₂  moles reacted = 64.0 g / 26 g/mol = 2.46 mol.

according to a molar ratio of 2:2.

the number of H₂O moles formed = a number of C₂H₂  moles reacted.

Therefore the number of H₂O moles produced = 2.46 mol

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Balance the following equations: (c) H2(g)+I2(s)⟶HI(s)H2(g)+I2(s)⟶HI(s)
kenny6666 [7]

Answer: H_2(g)+I_2(g)\rightarrow 2HI(s)

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Thus in the reactants, there are 2 atoms of hydrogen and 2 atoms of iodine .Thus there has to be 2 atoms of hydrogen and 2 atoms of iodine in the product as well. Thus a coefficient of 2 is placed in front of HI.

The balanced chemical reaction is:

H_2(g)+I_2(g)\rightarrow 2HI(s)

6 0
3 years ago
I don’t really understand, if anybody can help I’ll really appreciate it ! Thank you.
Alex_Xolod [135]

Answer:

175

Explanation:

3 0
3 years ago
In neutralization reaction _______ and ________ is produced.
dezoksy [38]

Answer:

In neutralization reaction water and a salt is produced.

4 0
2 years ago
Read 2 more answers
What is the density of 0.50 grams of gaseous carbon stored under 1.5 atm of pressure at a temperature of -20.0 C?
Colt1911 [192]

Answer: The density of 0.50 grams of gaseous carbon stored under 1.50 atm of pressure at a temperature of -20.0 °C is 0.867 g/L.

Explanation:

  • d = m/V, where d is the density, m is the mass and V is the volume.
  • We have the mass m = 0.50 g, so we must get the volume V.
  • To get the volume of a gas, we apply the general gas law PV = nRT

P is the pressure in atm (P = 1.5 atm)

V is the volume in L (V = ??? L)

n is the number of moles in mole, n = m/Atomic mass, n = 0.50/12.0 = 0.416 mole.

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature in K (T(K) = T(°C) + 273 = -20.0 + 273 = 253 K).

  • Then, V = nRT/P = (0.416 mol)(0.082 L.atm/mol.K)(253 K) / (1.5 atm) = 0.576 L.
  • Now, we can obtain the density; d = m/V = (0.50 g) / (0.576 L) = 0.867 g/L.
6 0
3 years ago
A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it r
Alchen [17]

6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

P\cdot V = n\cdot R\cdot T,

where

  • P is the pressure of the gas,
  • V is the volume of the gas,
  • n is the number of gas particles in the gas,
  • R is the ideal gas constant, and
  • T is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume V of the gas. Rearrange the ideal gas equation for volume:

V = \dfrac{n \cdot R \cdot T}{P}.

Both the temperature of the gas, T, and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

  • T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K},
  • T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}.

The volume of the gas is proportional to its temperature if both n and P stay constant.

  • n won't change unless the balloon leaks, and
  • consider P to be constant, for calculations that include T.

V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}.

Now, keep the temperature at T_1 =258.65\;\text{K} and change the pressure on the gas:

  • P_1 = 0.995\;\text{atm},
  • P_2 = 0.720\;\text{atm}.

The volume of the gas is proportional to the reciprocal of its absolute temperature \dfrac{1}{T} if both n and T stays constant. In other words,

V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}

(3 sig. fig. as in the question.).

See if you get the same result if you hold T constant, change P, and then move on to change T.

6 0
2 years ago
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