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3 years ago

PLEASE HELP ME

Chemistry

1 answer:

Flauer [41]3 years ago

7 0

Answer:

Explanation:

Li(s)+1/2 F₂(g)→LiF(s)

In this reaction one mole of product is formed from gaseous state of reactants . So this reaction will represent the reaction for which ΔH∘rxn equals to ΔH∘f .

In other reaction one mole is not produced if we balance the reaction. The reactant must be in gaseous state . ΔH∘f . represents heat of formation . ΔH∘rxn represents heat of reaction .

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What is the mass in grams of BaCl2 that is needed to prepare 200 mL of a 0.500 M solution

DENIUS [597]

Answer:

= 20.82 g of BaCl2

Explanation:

Given,

Volume = 200 mL

Molarity = 0.500 M

Therefore;

Moles = molarity × volume

= 0.2 L × 0.5 M

= 0.1 mole

But; molar mass of BaCl2 is 208.236 g/mole

Therefore; 0.1 mole of BaCl2 will be equivalent to;

= 208.236 g/mol x 0.1 mol

= 20.82 g

Therefore, the mass of BaCl2 in grams required is 20.82 g

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Which type of decay has the greatest mass? A. alpha B. beta gamma D. nuclear

coldgirl [10]

Answer:

Alpha has the greatest mass- A.

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3 years ago

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2 gallons is equal to how many liters

Arte-miy333 [17]

Answer:

7.57082

Explanation:

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3 years ago

An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg

lianna [129]

An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg of sodium phenate. What is the ph of the solution. The pka of phenol = 9.98. Mw of phenol and sodium phenate are 94 g/mol and 116 g/mol, respectively.

Volume of solution = 55 gallons = 208.2 L [ 1 gallon = 3.78 L]

moles of phenol = mass / molar mass = 4000 g / 94 = 42.55 moles

moles of sodium phenate = mass / molar mass = 1500 / 116 = 12.93 moles

pKa of phenol = 9.98

We know that the pH of buffer is calculated using Hendersen Hassalbalch's equation

pH = pKa + log [salt] / [acid]

volume is same for both the sodium phenate and phenol has we can directly take the moles of each in the formula

pH = 9.98 + log [12.93 / 42.55] = 9.46

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3 years ago

Unit: Chemical Quantities

Vaselesa [24]

Answer:

(See explanation for further details)

Explanation:

1) The quantity of moles of sulfur is:

$n = \frac{1.20\times 10^{24}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }$

$n = 1.993\,moles$

2) The number of atoms of helium is:

$x = (1.5\,moles)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)$

$x = 9.033\times 10^{23}\,atoms$

3) The quantity of moles of carbon monoxide is:

$n = \frac{4.15\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

$n = 0.689\,moles$

4) The number of molecules of sulfur dioxide is:

$x = (2.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 1.355\times 10^{24}\,molecules$

5) The quantity of moles of sodium chloride is:

$n = \frac{2.4\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

$n = 0.399\,moles$

6) The number of formula units of magnesium iodide is:

$x = (1.8\,moles)\cdot \left(6.022\times 10^{23}\,\frac{f.u.}{mole} \right)$

$x = 1.084\times 10^{24}\,f.u.$

7) The quantity of moles of potassium permanganate is:

$n = \frac{3.67\times 10^{23}\,f.u.}{6.022\times 10^{23}\,\frac{f.u.}{mol} }$

$n = 1.214\,moles$

8) The number of molecules of carbon tetrachloride is:

$x = (0.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 1.506\times 10^{23}\,molecules$

9) The quantity of moles of aluminium is:

$n = \frac{3.67\times 10^{23}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }$

$n = 0.609\,moles$

10) The number of molecules of oxygen difluoride is:

$x = (3.52\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 2.120\times 10^{24}\,molecules$

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3 years ago