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3 years ago

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A chemical company makes two brands of antifreeze. the first brand is 70% pure antifreeze, and the second brand is 95% pure anti

freeze. in order to obtain 160 gallons of a mixture that contains 80% pure antifreeze, how many gallons of each brand of antifreeze must be used?

1 answer:

Irina18 [472]3 years ago

6 0

64 gallons of 95%
96 gallons of 70%

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The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac

creativ13 [48]

Answer:

a) After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$ c) No $H_{2}$ was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$

c) $H_{2}$ was not the limiting reactant based on the mol to mol ratio of $H_{2}$ and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of $H_{2}$ would also be produced.

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3 years ago

Problem PageQuestion Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume

IgorC [24]

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 3.16 L of chlorine were consumed at STP.

Be sure your answer has the correct number of significant digits.

Answer: Thus volume of carbon tetrachloride that would be produced is 0.788 L

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = 3.16 L

n = number of moles = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $273K=$

$n=\frac{PV}{RT}$

$n=\frac{1atm\times 3.16L}{0.0820 L atm/K mol\times 273K}=0.141moles$

$CH_4+4Cl_2\rightarrow 4HCl+CCl_4$

According to stoichiometry:

4 moles of chlorine produces = 1 mole of carbon tetrachloride

Thus 0.141 moles of methane produces = $\frac{1}{4}\times 0.141=0.0352$ moles of carbon tetrachloride

volume of carbon tetrachloride = $moles\times {\text {Molar volume}}=0.0352mol\times 22.4L/mol=0.788L$

Thus volume of carbon tetrachloride that would be produced is 0.788 L

8 0

3 years ago

How many liters of hydrogen are required to react completely with 2.4L of oxygen to form water? 2H2 + O2 --> 2H2O

RoseWind [281]

Answer:

2.4 mole of oxygen will react with 2.4 moles of hydrogen

Explanation:

As we know

1 liter = 1000 grams

2H2 + O2 --> 2H2O

Weight of H2 molecule = 2.016 g/mol

Weight of water = 18.01 gram /l

2 mole of oxygen react with 2 mole of H2

2.4 mole of oxygen will react with 2.4 moles of hydrogen

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3 years ago

You leave gasoline in an open can and it vaporizes. What type of change is this?

UNO [17]

Answer:evaporation

Explanation:

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3 years ago

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A certain radioactive isotope decays at a rate of 0.2% annually. Determine the half-life of this isotope, to the nearest year

pychu [463]

Answer:

The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$

Where:

$m$ - Current isotope mass, measured in kilograms.

$t$ - Time, measured in years.

$\tau$ - Time constant, measured in years.

The solution of this differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

$\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%$

$1 - \frac{m(t+1)}{m(t)} = 0.002$

$1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002$

$1 - e^{-\frac{1}{\tau} } = 0.002$

$e^{-\frac{1}{\tau} } = 0.998$

$-\frac{1}{\tau} = \ln 0.998$

The time constant associated to the decay is:

$\tau = -\frac{1}{\ln 0.998}$

$\tau \approx 499.500\,years$

Finally, the half-life of the isotope as a function of time constant is given by the expression described below:

$t_{1/2} = \tau \cdot \ln 2$

If $\tau \approx 499.500\,years$ , the half-life of the isotope is:

$t_{1/2} = (499.500\,years)\cdot \ln 2$

$t_{1/2}\approx 346.227\,years$

The half-life of the radioactive isotope is 346 years.

6 0

3 years ago