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3 years ago

7

Relate what you've learned about the cell to how we inherit traits from our parents. Include the following vocab words, : nucleu

s, protein, DNA, Ribosome, characteristics, genes, and meiosis.: ( you must include all words ) MINIMUN 6 SENTENCES MUST BE WELL DEVELOPED THANKSS ! <3

2 answers:

Veronika [31]3 years ago

5 0

Meiosis makes sperm cells and eggs in our parents.
Genes are passed on from them to us.
As Genes carry characteristic, we inherit some characters of our either parent.
That Genetic code resides in DNA [ in it's nitrogenous bases ]
DNA resides under the nucleus of our cells.
When, it translated, it forms into "Protein"

[ We used all 6 words, but in sequence, with proper functioning/place ]

Hope this helps!

Anna [14]3 years ago

4 0

Answer:

thank you so much!!

Explanation:

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Answer:

$210.7~g~MgCO_3$

Explanation:

We have to start with the <u>reaction</u>:

$MgCO_3~->~MgO~+~CO_2$

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the <u>molar mass</u> of carbon dioxide:

$(12*1)+(16*2)=44~g/mol$

In other words: $1~mol~CO_2=~44~g~CO_2$ . With this in mind, we can calculate the moles:

$110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2$

Now, the <u>molar ratio</u> between carbon dioxide and magnesium carbonate is 1:1, so:

$2.5~mol~CO_2=2.5~mol~MgCO_3$

With the molar mass of $MgCO_3$ ( $(23.3*1)+(12*1)+(16*3)=84.3~g/mol$ . With this in mind, we can calculate the <u>grams of magnesium carbonate</u>:

$2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3$

I hope it helps!

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A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).

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First empirical formula of binary ionic compound is written between $Zn^{2+} (Positive ion)and F^{-} (Negative ion)$

First Formula would be $ZnF_{2}$

Second empirical formula is between $Zn^{2+}(Positive ion) and O^{2-}(Negative ion)$

Second Formula would be $Zn_{2}O_{2}$

Note : When the subscript are same they get cancel out, so $Zn_{2}O_{2}$ would be written as $ZnO$

Third empirical formula is between $Ni^{4+}(Positive ion) and F^{-}(Negative ion)$

Third Formula would be : $NiF_{4}$

Forth empirical formula is between $Ni^{4+}(Positive ion)and O^{2-}(negative ion)$

Forth Formula would be : $Ni_{2}O_{4}$ or $NiO_{2}$

Note- The subscript will be simplified and the formula will be written as $NiO_{2}$ .

The empirical formula of four binary ionic compounds are : $ZnF_{2}, ZnO, NiF_{4},NiO_{2}$

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