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Lemur [1.5K]
4 years ago
15

What is the subscript present in NaCl?

Chemistry
2 answers:
Marat540 [252]4 years ago
5 0

Answer:

NaCl, or sodium chloride, is commonly known as table salt. ... Subscripts show the ratio of atoms of different elements in the compound. Since there are four hydrogen atoms in the molecule, the subscript 4 follows the symbol for hydrogen.

Explanation:

jonny [76]4 years ago
3 0

Answer:

NaCl or sodium chloride, is commonly known as table salt. ... Subscripts show the ratio of atoms of different elements in the compound. Since there are four hydrogen atoms in the molecule, the subscript 4 follows the symbol for hydrogen.

Explanation:

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A 2.50-l flask contains a mixture of methane (ch4) and propane (c3h8) at a pressure of 1.45 atm and 20°c. when this gas mixture
Cerrena [4.2K]

Answer:- Mole fraction of methane in the original gas mixture is 0.854.

Solution:- From given volume, pressure and temperature, we could calculate the total moles of the gaseous mixture of methane and propane using ideal gas law as:

PV = nRT

n=\frac{PV}{RT}

V = 2.50 L

P = 1.45 atm

T = 20 + 273 = 293 K

Let's plug in the values in the equation:

n=(\frac{1.45*2.50}{0.0821*293})

n = 0.151

Let's say the solution has X moles of methane. Then moles of propane would be = (0.151 - X)

The combustion equations of methane and propane are:

CH_4+2O_2\rightarrow CO_2+2H_2O

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

From methane balanced equation, there is 1:1 mol ratio between methane and carbon dioxide. So, X moles of methane would produce X moles of carbon dioxide.

From balanced equation of propane, there is 1:3 mol ratio between propane and carbon dioxide. So, (0.151 - X) moles of propane would give 3(0.151 - X) moles of carbon dioxide.

So, total moles of carbon dioxide that we would get from methane and propane combustion are:

X + 3(0.151 - X)

From given data, 8.60 g of carbon dioxide are formed by the combustion of gas mixture.

moles of Carbon dioxide = 8.60g(\frac{1mol}{44g})

moles of carbon dioxide = 0.195 mol

Hence, 0.195 = X + 3(0.151 - X)

Let's solve this for X as:

0.195 = X + 0.453 - 3X

0.195 = 0.453 - 2X

2X = 0.453 - 0.195

2X = 0.258

X=\frac{0.258}{2}

X = 0.129

So, there are 0.129 moles of methane in the mixture.

moles of propane = 0.151 - 0.129 = 0.022

mole fraction of methane = \frac{moles of Methane}{total moles}

mole fraction of methane = \frac{0.129}{0.151}

mole fraction of methane = 0.854

Hence, the mole fraction of methane gas in the original gas mixture is 0.854.

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7. What volume of a 0.10 mol/l HCl solution is needed to neutralize 10 ml of a 0.15
svp [43]

Answer: The volume of a 0.10 mol/l HCl solution needed to neutralize 10 ml of a 0.15 mol/l LiOH solution is 15 ml

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is LiOH.

We are given:

n_1=1\\M_1=0.10mol/L\\V_1=?mL\\n_2=1\\M_2=0.15mol/L\\V_2=10mL

Putting values in above equation, we get:

1\times 0.10\times V_1=1\times 0.15\times 10\\\\V_1=15mL

Thus the volume of a 0.10 mol/l HCl solution needed to neutralize 10 ml of a 0.15 mol/l LiOH solution is 15 ml

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