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irina [24]
3 years ago
13

At birth, mammal babies are:

Chemistry
1 answer:
Lena [83]3 years ago
5 0

Answer:

It's A: Weak & Vulnerable

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Workout the formula for:<br>Magnesium Bromide <br>Potassium Chloride ​
madam [21]
Magnesium bromide= MgBr2
Potassium chloride= KCl
8 0
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Complete the chemical equation related to the formation of sodium chloride​
son4ous [18]

Answer:

sodium + chlorine --> sodium chloride

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Which of these factors are experienced by a space shuttle in orbit? Check all that apply.
Minchanka [31]
A. gravity (of any planet/star/celestial body around )

D. Inertia

E. Centripetal force
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3 years ago
Read 2 more answers
Write the balanced equation for the reaction between HCl and NaOH. Then, calculate the molarity of the NaOH.
andre [41]

Answer:

HCl (aq) + NaOH (aq) --> H2O (L) + NaCl (aq)

Explanation:

HCl is a strong acid while NaOH is a strong alkali. Hence both should dissociate completely in water and 1 mol of HCl will fully neutralise 1 mol of NaOH.

I'm assuming that Part 2 on molarity is part of a data based question that requires you to calculate the number of moles of NaOH based on the data provided and the equation that you are required to balance. Hence, I can't help you with it as I do not have the values.

5 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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