Someone Please Help Me out

can anyone help with this mole to mole/mole to grams question? any help would be greatly appreciated! Thanks!

goldenfox [79]

Answer:

Explanation:

a) 3.00 mol Cu x 2 mol Ag/ 1 mol Cu = 6 mol Ag

b) 3.00 mol Cu x 2 mol Ag/1 mol Cu x 107.87g Ag/ 1 mol Ag = 647.22g Ag

c) 3.00 mol Cu x 1 mol Cu(NO3)2/ 1 mol Cu = 3 mol Cu(NO3)2

d) 3.00 mol Cu x 1 mol Cu(NO3)2/ 1 mol Cu x 187.54g/ 1 mol Cu(NO3)2 =562.644g Cu(NO3)2

8 0

3 years ago

By titration, 15.0 mLmL of 0.1008 MM sodium hydroxide is needed to neutralize a 0.2053-gg sample of an organic acid. What is the

andre [41]

The question is incomplete, here is the complete question:

By titration, 15.0 mL of 0.1008 M sodium hydroxide is needed to neutralize a 0.2053-g sample of an organic acid. What is the molar mass of the acid if it is monoprotic.

Answer: The molar mass of monoprotic acid is 135.96 g/mol

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

For sodium hydroxide:

Molarity of NaOH solution = 0.1008 M

Volume of solution = 15.0 mL

Putting values in above equation, we get:

$0.1008M=\frac{\text{Moles of NaOH}\times 1000}{15.0}\\\\\text{Moles of NaOH}=\frac{(0.1008\times 15.0)}{1000}=0.00151mol$

As, the acid is monoprotic, it contains 1 hydrogen ion

1 mole of $OH^-$ ion of NaOH neutralizes 1 mole of $H^+$ ion of monoprotic acid

So, 0.00151 moles of $OH^-$ ion of NaOH will neutralize $\frac{1}{1}\times 0.00151=0.00151mol$ of $H^+$ ion of monoprotic acid

In monoprotic acid:

1 mole of $H^+$ ion is released by 1 mole of monoprotic acid

So, 0.00151 moles of $H^+$ ion will be released by $\frac{1}{1}\times 0.00151=0.00151mol$ of monoprotic acid

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of monoprotic acid = 0.00151 mole

Given mass of monoprotic acid = 0.2053 g

Putting values in above equation, we get:

$0.00151mol=\frac{0.2053g}{\text{Molar mass of monoprotic acid}}\\\\\text{Molar mass of monoprotic acid}=\frac{0.2053g}{0.00151mol}=135.96g/mol$

Hence, the molar mass of monoprotic acid is 135.96 g/mol

8 0

3 years ago

Which term is same for one mole of oxygen gas and one mole of water

miv72 [106K]

The number of atoms in one mole is same in both which is 6 x 10^23 ^23 means power 23

6 0

3 years ago

¿Cuáles son las características de los materiales de laboratorio? Por ejemplo: exactitud, resistencia a la temperatura, etc.

irina [24]

Answer:

gfchjklnmbvh

Explanation:

4 0

2 years ago

Read 2 more answers

Calnexin and calreticulin catalyze the removal of the final glucose residue from glycoproteins during the folding process. True

zloy xaker [14]

Answer:

Explanation:

A) False.

Glucosidase (not calnexin nor calreticulin) helps to remove glucose residue.

Both calnexin and calreticulin rather have an affinity for last glucose residue of misfolded protein (Only misfolded proteins are marked by glycosyltransferase by attaching glucose residue). They attach with misfolded protein and with the help of other proteins like ERp57 (a type of protein disulfide isomerase) and try to fold it properly. If protein is properly folded then glucosidase removes the glucose residue thereby releasing the properly folded protein from calnexin or calreticulin. and now protein is transported to the Golgi body. If folding is still not proper then the same cycle of glycosylation -binding of calnexin/calreticulin and effort to fold it properly is repeated.

B) True.

Transketolase is a key enzyme of the pentose phosphate pathway. It contains thiamine diphosphate (TPP) as a cofactor. it does transfer 2 carbon residue from a ketose to aldose. So, effectively it converts one ketose sugar to aldose with 2 carbonless and aldose to ketose with 2 carbon more.

C) True.

Theoretically, for the evolution of one molecule of oxygen, only 8 photons are required. But in practice, it is known that there are many variants like wavelength and the energy of the photon. The larger the wavelength, like the one which is used in PS1 (more than 700nM), the lesser the energy. Secondly, the energy of the photon is also wasted as heat energy. Because of these factors, more than 8 photons are needed in reality.

D) Wrong.

Fructose 2,6 bisphosphate is a key substrate and affects both the enzymes- phosphofructokinase and fructose bisphosphatase allosterically during gluconeogenesis. It strongly favors the breakdown of glucose during glycolysis by activating phosphofructokinase but it inhibits fructose bisphosphatase. Hence it activates the kinase enzyme while inhibiting the phosphatase and maintains a huge supply of glucose in the system.

E) Wrong.

The Calvin cycle shares similarity with the pentose phosphate pathway as both are involved in the synthesis of sugar (Triose and Ribose). However, it does not share similarity with enzymes of glycolysis (which is primarily focused on the breakdown of glucose) and gluconeogenesis.

8 0

3 years ago