5. How many moles of gas are in a 7.5 L container, at 1 atm of pressure, at 100 degrees

A gas evolved during the fermentation of sugar was collected at 22.5°C and 702 mmHg. After purification its volume was found to

miskamm [114]

Answer:

A) 0.95 mol

Explanation:

We will assume the gas given off in the fermentation is an ideal gas because that allows us to use the ideal gas equation.

PV = nRT

First let's convert all measurements to units that we can use

P = 702 mmHg * 1 atm/760 mmHg = 0.92368 atm

V = 25.0 L

R = 0.08206 L-atm/mol-K

T = 22.5 °C +273.15 = 295.65 K

PV = nRT

0.92368 atm * 25.0 L = n * 0.08206 L-atm/mol-K * 295.65 K

n = 0.9518 mol

4 0

3 years ago

What product has almost the same compounds as baking soda 10pts

mamaluj [8]

Ummm i actually dont know lol i swear i knew im sorry man or women ( im not trying to assume your gender

4 0

3 years ago

How is 0.246¯¯¯¯ written as a fraction in simplest form?

KiRa [710]

$0.246 = \dfrac{246}{1000} = \dfrac{246 \div 2}{1000 \div 2} = \dfrac{123}{500}$

4 0

3 years ago

When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.

Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

$Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3$

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

$0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2$

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

$0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2$

And the resulting percent yield:

$Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%$

Regards!

4 0

3 years ago

A 0. 462 g sample of a monoprotic acid is dissolved in water and titrated with 0. 180 m koh. what is the molar mass of the acid

Len [333]

The molar mass of the acid if 28. 5 ml of the koh solution is required to neutralize the sample is 90.23g/mol.

<h3>For the calculation of molarity of solution</h3>

Molarity = (moles of solute/volume of solution) × 1000

Given,

Molarity of KOH solution = 0.180 M

Volume of solution = 28.5 mL

0.180 = (moles of KOH/ 28.5) × 1000

Moles = (0.18× 28.5)/1000

= 0.00513 mol

<h3>Chemical equation for the reaction</h3>

HA + KOH ------- KA + H2O

1 moles of KOH reacts with 1 moles of HA.

So, 0.00513 moles of KOH react with 0.00513 moles of HA.

<h3>To calculate the molar mass for given number of moles</h3>

Number of moles= given mass/ Molar mass

Given,

Mass of HA = 0.462 g

Moles of HA = 0.00512 mol

0.00512 = 0.462/ Molar mass

Molar mass = 90.23 g/ mol.

Thus the molar mass of HA required to neutralize the 28.5 mL of KOH is 90.23g/mol.

learn more about molar mass or moles:

brainly.com/question/26416088

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3 0

1 year ago