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Svet_ta [14]
4 years ago
6

In balancing the nuclear reaction 7535Br → E + 01e, the identity of element E is ________. In balancing the nuclear reaction Br

E + e, the identity of element E is ________. Se U Kr Br none of the above
Chemistry
1 answer:
nordsb [41]4 years ago
4 0

<u>Answer:</u> The element E is Krypton (Kr).

<u>Explanation:</u>

In a nuclear reaction, the total mass and total atomic number remains the same.

The given nuclear reaction follows:

_{35}^{75}\textrm{Br}\rightarrow ^{A}_{Z}\textrm{X}+^0_{-1}\textrm{e}

  • <u>To calculate A:</u>

Total mass on reactant side = total mass on product side

75 = A + 0

A = 75

  • <u>To calculate Z:</u>

Total atomic number on reactant side = total atomic number on product side

35 = Z + (-1)

Z = 36

The isotopic symbol of element E is _{36}^{75}\textrm{Kr}

Hence, the element E is Krypton (Kr).

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What is the empirical formula for a compound that is 83.7% carbon and 16.3% hydrogen?
zubka84 [21]
Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100 g solution)

C: 83.7% = 83,7 g 
H: 16.3% = 16.3 g 

Let us use the above mentioned data (in g) and values will be ​​converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

C:  \dfrac{83.7\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 6.975\:mol

H: \dfrac{16.3\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 16.3\:mol

We note that the values ​​found above are not integers, so let's divide these values ​​by the smallest of them, so that the proportion is not changed, let's see:

C:  \dfrac{6.975}{6.975} = 1

H:  \dfrac{16.3}{6.975} \approx 2.3

Note: So the ratio in the smallest whole numbers of carbon to hydrogen is 3:7, t<span>hus, the minimum or empirical formula found for the compound will be:
</span>
\boxed{\boxed{C_3H_7}}\end{array}}\qquad\checkmark

I hope this helps. =)
8 0
3 years ago
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lyudmila [28]
C — all others endanger you.
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3 years ago
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How many grams of O2 are present in 44.1 L of O2 at STP?
ycow [4]

Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

First of all, the STP conditions refer to the standard temperature and pressure, where the values ​​used are: pressure at 1 atmosphere and temperature at 0°C. These values ​​are reference values ​​for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where:

  • P is the gas pressure.
  • V is the volume that occupies.
  • T is its temperature.
  • R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
  • n is the number of moles of the gas.

Then, in this case:

  • P= 1 atm
  • V= 44.1 L
  • n= ?
  • R= 0.082 \frac{atmL}{molK}
  • T= 0°C =273 K

Replacing in the expression for the ideal gas law:

1 atm× 44.1 L= n× 0.082 \frac{atmL}{molK}× 273 K

Solving:

n=\frac{1 atm x44.1 L}{0.082\frac{atmL}{molK}x273K}

n=1.97 moles

Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:

1.97 molesx\frac{32 g}{1 mole}= 63.04 g ≈ <u><em>63 g</em></u>

Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

Learn more about the ideal gas law:

  • <u>brainly.com/question/4147359?referrer=searchResults</u>
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A competent engineer would already know what has been done in the past to clean up oceanic oil residue.

Identifying needs and evaluating benefits and challenges are <em>value judgements</em> that are more appropriate for other decision-makers such as politicians.

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The outcome of that will be 44
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