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3 years ago

Convert .035 gallons to liters

Chemistry

2 answers:

kvasek [131]3 years ago

8 0

.035 gallons equals 0.1324894 L

taurus [48]3 years ago

4 0

I am sure that 0.035 gallons is equivalent to 0.13248941 liters,

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A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$

$60mL=0.06L$

Now we can calculate the concentration:

$M=\frac{0.00423mol}{0.06L}$

$M=0.0705$

Now, we can calculate the pOH (to calculate the pH), so:

$pOH=-Log(0.0705)$

$pOH=1.15$

Now we can calculate the pH value:

$14=~pH~+~pOH$

$pH=14-1.15=12.84$

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How many liters of CO2 are formed when 14.0 g of CaCO3 react at 1.00 atm and 1000K

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Explanation:

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Without consulting Appendix B, arrange each group in order of increasing standard molar entropy (S°). Explain.(c) SF₆(g), SF₄(g)

Andre45 [30]

The increasing order of standard molar entropy (S°) is as follow:

SF₄(g) < SF₆(g) < S₂F₁₀(g)

<h3>What is Entropy? </h3>

Entropy is defined as the randomness of the particle. It depends on temperature and pressure or number of particle per unit volume.

It is directly proportional to the temperature and pressure of the gas.

<h3>What is Standard Molar Entropy? </h3>

The standard molar entropy is defined as the entropy content of the one mole of pure substance at the standard state of temperature and pressure of interest.

The standard molar entropy is also defined as the total amount of entropy which 1 mole of the substance acquire, as it is brought from 0K to standard conditions of temperature and pressure.

The standard molar entropy depends on the molas mass of atom, molecules or compound.

SF₄(g) has lower standard molar entropy. Due to less complexity of this molecules.

While, complexity increases from SF₆(g) to S₂F₁₀(g). Therefore, the standard molar entropy of S₂F₁₀(g) is greater than SF₆(g).

Thus, we concluded that the increasing order of standard molar entropy (S°) is as follow:

SF₄(g) < SF₆(g) < S₂F₁₀(g)

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