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3 years ago

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How many moles of atoms are contained in the following 2.25g of k

1 answer:

ZanzabumX [31]3 years ago

8 0

Atomic mass (K)= 39.1 amu

therefore:
1 mol (k)---------------------39.1 g
x------------------------------- 2.25 g

x=(1 mol * 2.25 g) / 39.1 g=0.05754....≈0.06 moles

Answer: 0.06 moles.

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Stochiometry .How many grams of CO2 is produced when excess CS2 reacts with 4 mols of O2? CS2+O2-SO2 Please balance equation.

Travka [436]

Mass of CO₂ produced : 58.67 g

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

CS₂ + 3O₂ -------> CO₂ + 2SO₂

mol of CO₂ based on mol of O₂ as a limiting reactant(CS₂ as an excess reactant)

From the equation, mol ratio of mol CO₂ : mol O₂ = 1 : 3, so mol CO₂ :

$\tt \dfrac{1}{3}\times 4=\dfrac{4}{3}$

mass CO₂ (MW= 44 g/mol) :

$\tt \dfrac{4}{3}\times 44=58.67~g$

4 0

3 years ago

A sample of the chiral molecule limonene is 62% enantiopure. what percentage of each enantiomer is present? what is the percent

ikadub [295]

The mixture contains 62 % one isomer and 38 % the enantiomer.

Let’s say that the mixture contains 62 % of the (<em>R</em>)-isomer.

Then % (<em>S</em>) = 100 % -62 % = 38 %

ee = % (<em>R</em>) - % (<em>S</em>) = 62 % -38 % = 24 %

6 0

3 years ago

Read 2 more answers

Compute the number of grams(g) of lead in 22.5 moles of lead.

iogann1982 [59]

Answer:

4662 in grams

Explanation:

hope this helps :).

7 0

2 years ago

A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre

givi [52]

Answer:

$\mathbf{0.02 M}$

Explanation:

$\text{So, from the given question:}$

$\text{EDTA will make complex with}$ $V^{+3}$ $\text{and the remaining EDTA will react with }$ $Ga^{+3}$

$\text{Hence, the total concentration of}$ $V^{+3}$ & $Ga^{+3}$ $\text{will be equivalent to EDTA concentration.}$

$V_{EDTA} = 25 \ mL$

$V_{V^{+3}} = 59.0 \ mL$

$V_{Ga^{+3}} = 13.0 \ mL$

$M_{EDTA} = 0.0680 \ M$

$M_{V^{+3}} = ???(unknown)$

$M_{Ga^{+3}} = 0.0400 \ M$

$V^{+3} + EDTA \to V[EDTA] + EDTA(Excess) \to^{CoA} \ Ga[EDTA] _{complex}$

$M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})$

$0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18$

$x = \dfrac{1.18}{59}$

$\mathbf{x =0.02 \ M }$

5 0

3 years ago

Show work, thanks

GarryVolchara [31]

Answer:

Q.1

Given-

Volume of solution-1 L

Molarity of solution -6M

to find gms of AgNO3-?

Molarity = number of moles of solute/volume of solution in litre

number of moles of solute = 6×1= 6moles

one moles of AgNO3 weighs 169.87 g

so mass of 6 moles of AgNO3 = 169.87×6=1019.22

so you need 1019.22 g of AgNO3 to make 1.0 L of a 6.0 M solution

7 0

2 years ago