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Sladkaya [172]
3 years ago
6

How many L of 3.0 M H2SO4 solution can be prepared by using 100.0 mL OF 18 M H2SO4?

Chemistry
1 answer:
max2010maxim [7]3 years ago
6 0

Answer: A volume of 600 mL of 3.0 M H_{2}SO_{4} solution can be prepared by using 100.0 mL OF 18 M H_{2}SO_{4}.

Explanation:

Given: V_{1} = ?,        M_{1} = 3.0 M\\

V_{2} = 100.0 mL,       M_{2} = 18 M

Formula used to calculate the volume is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\3.0 M \times V_{1} = 18 M \times 100.0 mL\\V_{1} = \frac{18 M \times 100.0 mL}{3.0M}\\= 600 mL

Thus, we can conclude that a volume of 600 mL of 3.0 M H_{2}SO_{4} solution can be prepared by using 100.0 mL OF 18 M H_{2}SO_{4}.

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3 years ago
If 1.76 g of an ideal gas occupy 1.0 L at standard temperature and pressure (STP), what is the molar mass of the gas?
ycow [4]

Answer:

Explanation:

Whenever you see molar masses in gas law questions, more often than not density will be involved. This question is no different. To solve this, however, we will first need to play with the combined ideal gas equation PV=nRT to make it work for density and molar mass. The derivation is simple but for the sake of time and space, I will skip it. Hence, just take my word for it that you will end up with the equation:M=dRTPM = molar mass (g/mol)d = density (g/L)R = Ideal Gas Constant (≈0.0821atm⋅Lmol⋅K) T = Temperature (In Kelvin) P = Pressure (atm)As an aside, note that because calculations with this equation involve molar mass, this is the only variation of the ideal gas law in which the identity of the gas plays a role in your calculations. Just something to take note of. Back to the problem: Now, looking back at what we're given, we will need to make some unit conversions to ensure everything matches the dimensions required by the equation:T=35oC+273.15= 308.15 KV=300mL⋅1000mL1L= 0.300 LP=789mmHg⋅1atm760mmHg= 1.038 atmSo, we have almost everything we need to simply plug into the equation. The last thing we need is density. How do we find density? Notice we're given the mass of the sample (0.622 g). All we need to do is divide this by volume, and we have density:d=0.622g0.300L= 2.073 g/LNow, we can plug in everything. When you punch the numbers into your calculator, however, make sure you use the stored values you got from the actual conversions, and not the rounded ones. This will help you ensure accuracy.M=dRTP=(2.073)(0.0821)(308.15)1.038= 51 g/molRounded to 2 significant figuresNow if you were asked to identify which element this is based on your calculation, your best bet would probably be Vandium (molar mass 50.94 g/mol). Hope that helped :) 

8 0
2 years ago
How much faster does ethane (C2H6) gas travel compared to chlorine gas?
zimovet [89]

Answer:

\large \boxed{\text{54 $\, \%$ faster }}

Explanation:

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if temperature is constant.  

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if we are comparing two gases,

\dfrac{v_{2}}{v_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}

Let chlorine be Gas 1 and ethane be Gas 2

Data:

M₁ =  70.91 g/mol

M₂ = 30.07 g/mol

Calculation

\begin{array}{rcl}\dfrac{v_{2}}{v_{1}} & = & \sqrt{\dfrac{M_{1}}{M_{2}}}\\\\& = & \sqrt{\dfrac{70.91}{30.07}}\\\\& = & \sqrt{2.358}\\\\& = & \mathbf{1.54}\\\end{array}\\\text{Ethane molecules travel at 1.54 times the speed of chlorine molecules}\\\text{or $\large \boxed{\textbf{54 $\%$ faster }}$ than chlorine molecules}

5 0
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