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Serggg [28]
3 years ago
9

In an experiment, a student needs 250.0 mL of a 0.100 M copper (II) chloride solution. A stock solution of 2.00 M copper (II) ch

loride is available. How much of the stock solution is needed
Chemistry
1 answer:
LekaFEV [45]3 years ago
8 0

Answer : The volume of stock solution needed are, 12.5 mL

Explanation :

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the initial molarity and volume of copper (II) chloride.

M_2\text{ and }V_2 are the final molarity and volume of stock solution of copper (II) chloride.

We are given:

M_1=0.100M\\V_1=250.0mL\\M_2=2.00M\\V_2=?

Putting values in above equation, we get:

0.100M\times 250.0mL=2.00M\times V_2\\\\V_2=12.5mL

Hence, the volume of stock solution needed are, 12.5 mL

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katrin2010 [14]

Answer:

0.062mol

Explanation:

Using ideal gas law as follows;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821Latm/molK)

T = temperature (K)

Based on the information provided;

P = 152 Kpa = 152/101 = 1.50atm

V = 0.97L

n = ?

T = 12°C = 12 + 273 = 285K

Using PV = nRT

n = PV/RT

n = (1.5 × 0.97) ÷ (0.0821 × 285)

n = 1.455 ÷ 23.39

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3 years ago
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Morgarella [4.7K]

Answer:

Molec_{\ H_{tot}}=1.206x10^{25}molec

Explanation:

Hello.

In this case, taking into account that HCl has one molecule of hydrogen per mole of compound which weights 36.45 g/mol, we compute the number of molecules of hydrogen in hydrochloric acid by considering the given mass and the Avogadro's number:

molec_{\ H}=3.65gHCl*\frac{1molHCl}{36.45gHCl} *\frac{1molH}{1molHCl}*\frac{6.022x10^{23}molec_\ H}{1molH}  =6.03x10^{22}molec

Now, from the 180 g of water, we see two hydrogen molecules per molecule of water, thus, by also using the Avogadro's number we compute the molecules of hydrogen in water:

molec_{\ H}=180gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O}*\frac{6.022x10^{23}molec_\ H}{1molH}  =1.20x10^{25}molec

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Answer:

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Solving for Mass,

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As,

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Putting values in equation 1,

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Step 2: Convert Grams into Pounds;

As,

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So,

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