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2 years ago

3.01x10^15 atoms of S= g

Chemistry

1 answer:

astra-53 [7]2 years ago

4 0

There are 1.65 × 10^(-7) g S in 3.01 × 10^15 atoms S.

Step 1. Use Avogadro’s number to convert atoms of S to moles of S

Moles of S = 3.01 × 10^15 atoms S × (1 mol S/6.022 × 10^23 atoms S)

= 5.148 × 10^(-9) mol S

Step 2. Use the molar mass of S to convert moles of S to grams of S

Mass of S = 5.148 × 10^(-9) mol S × (32.06 g S/1 mol S) = 1.65 × 10^(-7) g S

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A chemist prepares a solution of barium acetate by measuring out of barium acetate into a volumetric flask and filling the flask

leonid [27]

The given question is incomplete. The complete question is :

A chemist prepares a solution of barium acetate by measuring out 32 g of barium acetate into a 350 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium acetate solution. Round your answer to significant digits.

Answer: The concentration of barium acetate solution is 0.375 mol/L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $Ba(CH_3COO)_2$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{32g}{255g/mol}=0.125mol$

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.125\times 1000}{350ml}$

$Molarity=0.357M$

Therefore, the concentration of solution is 0.375 mol/L

8 0

2 years ago

In glycolysis, if glucose is labeled at the carbon 6 position (see page 1 for numbering of carbons in glucose) A) the carbon wit

Oliga [24]

Answer:

D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.

Explanation:

Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.

(1) Preparatory phase

During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.

When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone phosphate (DHAP). In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.

(2) Pay-off phase

During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.

It is pertinent to mention here that BPG has a mixed anhydride and the bond at C1 is a very high energy bond. In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.

3 0

2 years ago

22.5 g of silver nitrate reacts with excess magnesium bromide, determine the mass

Setler [38]

Answer:

9.82 g of Mg(NO₃)₂

Explanation:

Let's determine the reaction:

2AgNO₃ + MgBr₂ → Mg(NO₃)₂ + 2AgBr

2 moles of nitrate silver reacts with MgBr₂ in order to produce 1 mol of magnesium nitrate and silver bromide.

We determine the moles of AgNO₃

22.5 g . 1mol / 169.87g = 0.132 moles

Ratio is 2:1.

2 moles of silver nitrate can produce 1 mol of magnesium nitrate

Then, our 0.132 moles may produce (0.132 . 1)/ 2 = 0.0662 moles

We convert moles to mass:

0.0662 mol . 148.3 g/ mol = 9.82 g

6 0

2 years ago

What is the relationship between the amount of substance ( concentration) and reaction rates?

jekas [21]

Answer:

Increasing the concentration of the reagents makes the collision between two molecules of the reagents more likely, thereby increasing the probability that the reaction will occur between these reagents.

As for the relationship between concentration and volume, density also comes into play, a higher volume, lower molarity and also lower concentration.

The pressure when increasing could generate a closer approach between the particles, therefore generating an increase in the reaction speed.

Pressure and volume are related but inversely proportional, therefore if the volume increases the pressure decreases and so on.

the reaction rate increases as the contact surface area increases. This is due to the fact that more solid particles are exposed and can be reached by reactant molecules.

A perfect reaction where the collision is promoted and the reaction speed advances is with the presence of a solvent, with an increase in pressure and a decrease in volume, with an increase in the exposure of the surface, with the presence of a catalyst, with increasing temperature and with increasing entrance

Explanation:

The reaction rate is defined as the amount of substance that is transformed into a certain reaction per unit of volume and time. For example, the oxidation of iron under atmospheric conditions is a slow reaction that can take many years but over time it is oxidized sooner or later by the oxygenation of its surface layer, but the combustion of butane in a fire is a reaction that happens in fractions of seconds, giving rise to an exothermic reaction with products such as CO2 and H2O

5 0

2 years ago

S8 + 24 F2 ⟶ 8 SF6

Arturiano [62]

Answer:

Theoretical Yield of SF₆ = 2.01 moles

Explanation: If you understand and can apply the methodology below, you will find it applies to ALL chemical reaction stoichiometry problems based on the balanced standard equation; i.e., balanced to smallest whole number coefficients.

Solution 1:

Rule => Convert given mass values to moles, solve problem using coefficient ratios. Finish by converting moles to the objective dimensions.

Given S₈ + 24F₂ => 8SF₆

425g 229g ?

= 425g/256g/mol. = 226g/38g/mol.

= 1.66 moles S₈ = 6.03 moles F₂ <= Limiting Reactant

Determining Limiting Reactant => Divide moles each reactant by their respective coefficient; the smaller value will always be the limiting reactant.

S₈ = 1.66/1 = 1.66

F₂ = 6.03/24 = 0.25 => F₂ is the limiting reactant

Determining Theoretical Yield:

Note: When working problem do not use the division ratio results for determining limiting reactant. Use the moles F₂ calculated from 229 grams F₂ => 6.03 moles F₂. The division procedure to define the smaller value and limiting reactant is just a quick way to find which reactant controls the extent of reaction.

Given S₈ + 24F₂ => 8SF₆

425g 229g ?

= 425g/256g/mol. = 226g/38g/mol.

= 1.66 moles S₈ = 6.03 moles F₂ <= Limiting Reactant

Max #moles SF₆ produced from 6.03 moles F₂ and an excess S₈

Since coefficient values represent moles, the reaction ratio for the above reaction is 24 moles F₂ to 8 moles SF₆. Such implies that the moles of SF₆ (theoretical) calculated from 6.03 moles of F₂ must be a number less than the 6.03 moles F₂ given. This can be calculated by using a ratio of equation coefficients between 24F₂ and 8SF₆ to make the outcome smaller than 6.03. That is,

moles SF₆ = 8/24 x 6.03 moles = 2.01 moles SF₆ (=> theoretical yield)

S₈ + 24F₂ => 8SF₆

moles SF₆ = 8/24(6.03) moles = 2.01 moles

You would NOT want to use 24/8(6.03) = 18.1 moles which is a value >> 6.03.

This analysis works for all reaction stoichiometry problems.

Convert to moles => divide by coefficients for LR => solve by mole mole ratios from balanced reaction and moles of given.

____________________

Here's another example just for grins ...

C₂H₆O + 3O₂ => 2CO₂ + 3H₂O

Given: 253g 307g ? ?

a. Determine Limiting Reactant

b. Determine mass in grams of CO₂ & H₂O produced

Limiting Reactant

moles C₂H₆O = 253g/46g/mol = 5.5 moles => 5.5/1 = 5.5

moles O₂ = 307g/32g/mol = 9.6 moles => 9.6/24 = 0.4 ∴ O₂ is L.R.

But the problem is worked using the mole values; NOT the number results used to ID the limiting reactant.

C₂H₆O + 3O₂ => 2CO₂ + 3H₂O

------------ 9.6 mole (L.R.) ? ?

mole yield CO₂ = 2/3(9.6)mole = 6.4 mole (CO₂ coefficient < O₂ coefficient)

mole yield H₂O = 9.6mole = 9.6mole (coefficients O₂ & CO₂ are same.)

mole used C₂H₆O = 1/3(9.6)mole = 3.2 mole (coefficient C₂H₆O < coefficient O₂)

For grams => moles x formula weight (g/mole)

7 0

2 years ago