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4 years ago

Fission and Fusion of Atomic Nuclei:

Chemistry

1 answer:

bogdanovich [222]4 years ago

6 0

Answer:

C B A D

Explanation:

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Recording station x has determined that the epicenter of a recent earthquake was 250 km away. no information is available yet fr

KatRina [158]

The epicenter was located somewhere on a circle centered at Recording station X, with a radius of 250 km.<span>
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3 years ago

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What is wrong with the way that the hypothesis, “Chocolate may cause pimples”, is written?

DaniilM [7]

Answer:

More sweet and cute with the little ones on the surface and a little more

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4 years ago

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An object has a mass of 5.52 g and a volume of 12 mL. What is the density of the object?

Rudik [331]

Answer:

<h2>Density = 0.46 g/mL</h2>

Explanation:

Density of a substance can be found by using the formula

<h3> $Density = \frac{mass}{volume}$ </h3>

From the question

mass = 5.52 g

volume = 12 mL

Substitute the values into the above formula and solve for the Density

That's

<h3> $Density = \frac{5.52}{12}$ </h3>

We have the final answer as

<h3>Density = 0.46 g/mL</h3>

Hope this helps you

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3 years ago

An experiment produced 0.10 g CO2, with a volume of 0.056 L at STP. If the accepted density of CO2 at STP is 1.96 g/L, what is t

Pachacha [2.7K]

The density of CO2 getting from experiment is 0.1/0.056 = 1.79 g/L. The percent error of this is (1.96 -1.79)/1.96*100%=8.67%. So the approximate percent error is 8.67%.

3 0

3 years ago

a water sample is found to have a cl- content of 100ppm as nacl what is the concentration of chloride in moles per liter

ladessa [460]

Answer:

The concentration of chloride ion is $2.82\times10^{-3}\;mol/L$

Explanation:

We know that 1 ppm is equal to 1 mg/L.

So, the $Cl^-$ content 100 ppm suggests the presence of 100 mg of $Cl^-$ in 1 L of solution.

The molar mass of $Cl^-$ is equal to the molar mass of Cl atom as the mass of the excess electron in $Cl^-$ is negligible as compared to the mass of Cl atom.

So, the molar mass of $Cl^-$ is 35.453 g/mol.

Number of moles = (Mass)/(Molar mass)

Hence, the number of moles (N) of $Cl^-$ present in 100 mg (0.100 g) of $Cl^-$ is calculated as shown below:

$N=\frac{0.100\;g}{35.453\;g/mol}=2.82\times 10^{-3}\;mol$

So, there is $2.82\times10^{-3}\;mol$ of $Cl^-$ present in 1 L of solution.

5 0

4 years ago