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3 years ago

If a penny is made of 3.11 grams of copper, how many atoms of copper are in the penny

Chemistry

1 answer:

Pie3 years ago

8 0

Answer:

2.94x10²² atoms of Cu

Explanation:

We must work with NA to solve this, where NA is the number of Avogadro, number of particles of 1 mol of anything.

Molar mass Cu = 63.55 g/mol

Mass / Molar mass = Mol → 3.11 g / 63.55 g/m = 0.0489 moles

1 mol of Cu has 6.02x10²³ atoms of Cu

0.0489 moles of Cu, will have (0.0489 .NA)/ 1 = 2.94x10²² atoms of Cu

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(a.) a 0.7549g sample of the compound burns in o2(g) to produce 1.9061g of co2(g) and 0.3370g of h2o(g).

Natali [406]

The individual mass of C, H and O in given sample are 0.5196 g, 0.0374 g and 0.1979 g respectively.

Moles of CO2 formed can be calculated as

= Mass of CO2 / Molar mass of CO2

= 1.9061 / 44 = 0.0433 moles

<h3>Calculation of no. of moles of carbon</h3>

Now, moles of C which is present in one mole of CO2 = 1 mole

Moles of C in 0.0433 moles of CO2 = 0.0433 moles

As we know that, molar mass of C = 12 g / mol

Mass of C in 0.7549 g of given sample can be calculated as

= 0.0433 × 12 =0.5196 g

Mass of H2O formed = 0.3370 g

Similarly, Molar Mass of H2O = 18 g / mol

Moles of H2O = 0.3370 / 18 = 0.0187 moles

Moles of H present in 1 mole of H2O = 2 moles

Moles of H present in 0.0187 mole of H2O = 2 × 0.0187 = 0.0374 moles

Molar mass of H = 1 g / mol

Mass of H contained in 0.7549 g of sample = 1 × 0.0374= 0.0374 g

Mass of O in 0.7549 g sample can be calculated as

= 0.7549 – [(Mass of C ) + (Mass of H) ]

= 0.7549 – [ (0.5196) + (0.0374) ]

= 0.1979 g

Thus, we calculated that the individual mass of C, H and O in given sample are 0.5196 g, 0.0374 g and 0.1979 g respectively.

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DISCLAIMER: THE above question is incomplete. Complete question is given below:

A 0.7549g sample of the compound burns in o2(g) to produce 1.9061g of co2(g) and 0.3370g of h2o(g). Calculate the individual mass of C, H and O in the given sample.

4 0

1 year ago

The correct sequence where reactivity towards oxygen increases.

ziro4ka [17]

Answer:

Option D is good to go!

Explanation: as per the reactivity series more reactive substances will react with the counterpart substance.The most reactive substance here is calcium while the least reactive is aluminium, the magnesium comes in between.As per their reactivity, these substances will react with oxygen.

Explanation:

3 0

3 years ago

Explain, in terms of electrons, why the bonding in strontium sulfide, SrS, is similar to the bonding in magnesium bromide, MgBr2

Aleksandr-060686 [28]

Answer: both compounds have ionic bond between metal and non-metal

Explanation: both Sr and Mg are earth alkaline metals and form ions Mg^2+

And Sr^2+. Br forms ion Br^- and S ion is S^2+.

5 0

3 years ago

An analytical chemist is titrating of a solution of nitrous acid with a solution of . The of nitrous acid is . Calculate the pH

Burka [1]

Answer:

pH = 2.69

Explanation:

The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

<em />

The reaction of HNO₂ with KOH is:

HNO₂ + KOH → NO₂⁻ + H₂O + K⁺

Moles of HNO₂ and KOH that react are:

HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>

KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>

That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:

NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻

HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂

It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:

pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]

pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]

8 0

3 years ago

Dehydrohalogenation of 1-chloro-1-methylcyclopropane affords two alkenes (A and B) as products.

UNO [17]

Explanation:

Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:

Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.

Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.

However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.

Elimination of the hydrogen from the methyl group is easier.

Thus, the major product will A

4 0

3 years ago