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Anna [14]
3 years ago
11

Given the following equation formats, which form matches a single replacement

Chemistry
1 answer:
Ber [7]3 years ago
7 0

Answer:

The answer is c. A + BC → AC + B

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The number of electrons in an atom's outermost valence shell governs its bonding behaviour. Therefore, we group elements whose atoms have the same number of valence electrons together in the Periodic Table. ... This tendency is called the octet rule, because the bonded atoms share eight valence electrons.
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What type of reaction is Ni + MgSo4
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Chemical- it produces ammonia.
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How can density be used to determine the identity of a pure substance?
OLEGan [10]

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Explanation:

You can identify an unknown substance by measuring its density and comparing your result to a list of known densities. Density = mass/volume. Assume that you have to identify an unknown metal.

5 0
3 years ago
Simplify the ratio 32:21 in its simplest form ..show your working​
makvit [3.9K]

Answer:

32/21

Explanation:

Find the GCD (or HCF) of numerator and denominator. GCD of 32 and 21 is 1.

32 ÷ 121 ÷ 1.

Reduced fraction: 3221. Therefore, 32/21 simplified to lowest terms is 32/21.

8 0
3 years ago
What is the energy released in the fission reaction 1 0 n + 235 92 U → 131 53 I + 89 39 Y + 16 1 0 n 0 1 n + 92 235 U → 53 131 I
mihalych1998 [28]

<u>Answer:</u> The energy released in the given nuclear reaction is 94.99 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

_{92}^{235}\textrm{U}+_0^1\textrm{n}\rightarrow _{53}^{131}\textrm{I}+_{39}^{89}\textrm{Y}+16_{0}^1\textrm{n}

We are given:

Mass of _{92}^{235}\textrm{U} = 235.043924 u

Mass of _{0}^{1}\textrm{n} = 1.008665 u

Mass of _{53}^{131}\textrm{I} = 130.9061246 u

Mass of _{39}^{89}\textrm{Y} = 88.9058483 u u

To calculate the mass defect, we use the equation:

\Delta m=\text{Mass of reactants}-\text{Mass of products}

Putting values in above equation, we get:

\Delta m=(m_U+m_{n})-(m_{I}+m_{Y}+16m_{n})\\\\\Delta m=(235.043924+1.008665)-(130.9061246+88.9058483+16(1.008665))=0.1019761u

To calculate the energy released, we use the equation:

E=\Delta mc^2\\E=(0.1019761u)\times c^2

E=(0.1019761u)\times (931.5MeV)  (Conversion factor:  1u=931.5MeV/c^2  )

E=94.99MeV

Hence, the energy released in the given nuclear reaction is 94.99 MeV.

3 0
3 years ago
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