1.94 grams of magnesium to moles

Answer:

Applied science is generally engineering, which develops technology, although there might be dialogue between basic science and applied science (research and development). Medical sciences such as medical microbiology are examples of applied sciences.

Explanation:

the balanced chemical equation for decomposition of HgO is as follows

2HgO --> 2Hg + O₂

stoichiometry of HgO to O₂ is 2:1

number of HgO moles heated are - 3.00 g / 216.59 g/mol = 0.0139 mol

according to stoichiometry of reaction -

number of O₂ moles formed = 0.0139 mol/ 2 = 0.00695 mol

mass of O₂ to be formed - 0.00695 mol x 32.00 g/mol = 0.2224 g

but the actual yield = 0.195 g

percent yield = actual yield / theoretical yield x 100 %

percent yield = 0.195 g / 0.2224 g x 100 % = 87.7 %

answer is 87.7 %

Answer:

The correct answer is 87.8%

The correct answer is 9.6h.

As you know, a radioactive isotope's nuclear half-life tells you exactly how much time must pass in order for an initial sample of this isotope to be halved.

Using the formula , A = Ao.

where , A- final mass after decay

Ao - initial mass

n - the number of half-lives that pass in the given period of time

Now, putting all the values, we get

1.3 × mg = 0.050 mg ×

Take the natural log of both sides of the equation to get,

㏑ = ㏑

㏑ = n. ln

n = 1.6

Since n represents the number of half-lives that pass in a given period of time, you can say that

t= 1.6 × 6 h

t = 9.6h
Hence, it will take 9.6 h  until the radioactive isotope decays.

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Final answer:

Using the formula for radioactive decay and the provided half-life of technetium-99m, it can be calculated that it takes approximately 28.5 hours for 0.050 mg of technetium-99m to decay to a quantity of 1.3 x 10^-2 mg.

Explanation:

The decay of a radioactive isotope is an exponential process based on the half-life, which is, in turn, constant for any given isotope. The general formula for the remaining quantity of a radioactive isotope after a given time is given by: N = N0 (0.5) ^(t/t1/2), where (N0) is the initial amount, (N) is the remaining amount, (t) is time, and (t1/2) is the half-life of the isotope. In this case, we are given the initial quantity (N0 = 0.050 mg), the remaining quantity (N = 1.3 x 10^-2 mg), and the half-life (t1/2 = 6.0 hours).

We can solve for time (t) in the equation: N = N0 (0.5) ^(t/t1/2). Plugging in the values, we get 1.3 x 10^-2 = 0.050 x (0.5)^(t/6), and solving for t, we find that it takes approximately 28.5 hours for the technetium-99m to decay to 1.3 x 10^-2 mg.

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Final answer:

In the reaction of hydrochloric acid and calcium carbonate, to balance the equation, the correct order of numbers would be 2 for HCl, and 1 for CaCO3, CaCl2, CO2, and H2O.

Explanation:

When hydrochloric acid (HCl) reacts with calcium carbonate (CaCO3) in concrete, it forms calcium chloride (CaCl2), carbon dioxide (CO2), and water (H2O). To balance this reaction, you could input the numbers: 2 HCl(aq) + 1 CaCO3(s) → 1 CaCl2(aq) + 1 CO2(g) + 1 H2O(l). This phenomenon is part of the larger concept known as a neutralisation reaction, where an acid reacts with a base to produce a salt and water. In this case, the hydrochloric acid is the acid and the calcium carbonate in the concrete is the base. This is also an example of a gas evolution reaction, as carbon dioxide gas is being produced.

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Final answer:

The molarity of HCl after the first dilution is 0.024 M and after the second dilition, it is 0.0036M.

Explanation:

The problem at hand is a solution dilution problem in the field of Chemistry, usually tackled by the use of formula M1V1=M2V2, where M1 and V1 are the original molarity and volume, and M2 and V2 are the molarity and volume after dilution. In the first dilution, applying this formula gives (0.15 M)(4.00 mL) = (M2)(25.00 mL), solving for M2 gives a value of 0.024 M. The second dilution would similarly have the equation (0.024 M)(7.5 mL) = (M3)(50.0 mL), which gives M3 approximately 0.0036 M. Hence, the molarity of the first dilution would be 0.024 M and the molarity of the second dilution would be 0.0036 M.

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