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VikaD [51]
3 years ago
11

How many moles of caffeine are in 130 mg of caffeine coffee?

Chemistry
1 answer:
Bond [772]3 years ago
4 0

Answer:

0.00067mols

Explanation:

firs convert 130mg to grams which is 0.13g, then find the molar mass of caffeine which is 194.19g/mol.

0.13g/194.19gmol=0.0067mols

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. an ionic compound is composed of 34.95 g of iron and 15.05 g of oxygen. find the empirical formula of this compound.
Contact [7]

Empirical formula of ionic compound is FeO. In which the composition of atoms is 1 : 1.

Empirical formula of an ionic compound is defined as the formula which gives whole number ratio of atoms of various elements present in molecule of compund.

mass of iron in compound = 34.95 g

molar mass of iron = 55.8 g

mass of oxygen in compound = 15.05 g

molar mass of oxygen = 32 g

number of moles of iron present in the compound are ratio of mass of iron in compound/ molar mass of iron

number of moles of iron in compound= 34.95 / 55.8 = 0.6263 ~ 1

number of moles oxygen in compound= 15.05/ 32 = 0.473 ~ 0.5

the ratio of the number of oxygen atoms to number of iron atoms present in one formula unit of iron compund is 2×0.5 / 1 = 1 : 1

Hence , the required empirical formula of iron compound is FeO.

To learn more about Emiprical formula, refer:

brainly.com/question/1439914

#SPJ4

3 0
1 year ago
Consider the reaction pathway graph below.
Alex17521 [72]
Hello!

The reaction that the graph represents is A. Exothermic because Hrxn=-167 kJ

To calculate Hrxn we apply the following equation:

Hrxn=Hproducts-Hreagents=-625kJ-(-458kJ)=-167kJ

Looking at the graph, and at the result of the calculations, we can see that the enthalpy of the products is lower than the enthalpy of the reagents, because the sign is negative. That means that the reaction releases energy in the form of heat and that the reaction is exothermic.

Have a nice day!
6 0
4 years ago
How many moles are there in 7.5 L Of H2
True [87]
So multiply number of moles x number of atoms/mole = 1.8066 x 10^24 atoms of H2. One mole of any gas at STP has a volume of 22.4 L. So first determine the number of moles of gas you have.
for example do 7
7.5 \div 22.4
that 's what I think
3 0
3 years ago
Which tool is used to measure mass?
kherson [118]

Answer: c. balance

Explanation: Mass is the amount of matter contained in a body.

3 0
3 years ago
Read 2 more answers
g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete
Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899        0.01714        0           0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714    0.01714 (end)

0.00185  | 0 → 0.01714    0.01714 (end)  

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

8 0
3 years ago
Read 2 more answers
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