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Solnce55 [7]
3 years ago
7

Definition for outcome variable

Chemistry
1 answer:
Deffense [45]3 years ago
8 0
The dependent variable is sometimes called "the outcome variable."
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Predictable moderate winds between 5 & 25 degrees North or South latitude are called?
Grace [21]

Answer:

احبـــــــــــــــــــك

لول ملل اقولج انا اسفه

6 0
3 years ago
Please help me!!!!!!​
Leona [35]

The answer is B

To write the equilibrium constant for an equation, all you have to do is divide the products by the reactants. The reactants are always on the left side, and the products are always on the right side. The coefficients of the elements will be written as the exponent of that same element. However, in this equation, we do not have to write any exponents, as there are no coefficient but 1.

5 0
3 years ago
For the following problem convert both the reactants to moles and balance chemical equationsThe reaction of 167 g Fe2O3 with 85.
saul85 [17]

Let's start by balancing the reaction:

Fe_2O_3+CO\longrightarrow Fe+CO_2

As we can see, C appears only on two comopunds, CO and CO₂, and since both have 1 C each, their coefficients have to be the same for C to be balanced. However, CO has 1 O and CO₂ has 2, so there is a difference of 1 O betwee them.

The other source of O is Fe₂O₃, that has 3 O. So, we must choose a coefficient for CO and CO₂ such that the difference between the numbers of O is a multiple of 3, that way we can fix this difference with the O from Fe₂O₃. So, we can put coefficients of 3 on both of them:

Fe_2O_3+3CO\longrightarrow Fe+3CO_2

That way, we maintained C balanced (3 on each side) and now we have 3 + 3 O on the left side and 6 O on the right side, so the same amount.

Now, we just have to calance Fe, but it is easy since we have it alone in Fe. Since we have 2 on the left side, it is enough to put a coefficient of 2 on Fe to get the balanced reaction:

Fe_2O_3+3CO\longrightarrow2Fe+3CO_2

Now, to convert from mass to number of moles, we need the molar masses of the reactants, which we can calculate from the atomic weights of the elemnts in each of them:

M_{Fe_2O_3}=2\cdot M_{Fe}+3\cdot M_O=(2\cdot55.845+3\cdot15.9994)g/mol=159.6882g/molM_{CO}=1\cdot M_C+1\cdot M_O=(1\cdot12.0107+1\cdot15.9994)g/mol=28.0101g/mol

Now, we can convert their masses to number of moles:

\begin{gathered} M_{Fe_{2}O_{3}}=\frac{m_{Fe_2O_3}}{n_{Fe_{2}O_{3}}} \\ n_{Fe_2O_3}=\frac{m_{Fe_2O_3}}{M_{Fe_{2}O_{3}}}=\frac{167g}{159.6882g/mol}=1.045787\ldots mol \end{gathered}\begin{gathered} M_{CO}=\frac{m_{CO}}{n_{CO}} \\ n_{CO}=\frac{m_{CO}}{M_{CO}}=\frac{85.8g}{28.0101g/mol}=3.063180\ldots mol \end{gathered}

Now, to determine the limiting reactant, we need to divide both the number of mole by their coefficients on the balanced reaction, so we can see how many we need per reaction of each:

\begin{gathered} Fe_2O_3\colon\frac{n_{Fe_2O_3}}{1}=\frac{1.045787\ldots mol}{1}=1.045787\ldots mol \\ CO\colon\frac{n_{CO}}{3}=\frac{3.063180\ldots mol}{3}=1.021060\ldots mol \end{gathered}

Now, the limiting reactant is the one we have less number of moles per reaction. We can see that we have less CO than Fe₂O₃, so the limiting reactant is CO.

4 0
1 year ago
Do you know the answer if so pls answer
igomit [66]

Answer:

They are both pretty soft for metals, but magnesium is significantly harder than calcium using this scale . Mg = 2.5, Ca = 1.75. The larger the number, the harder

Explanation:

6 0
2 years ago
Will mark brainliest​
MArishka [77]

Answer:

D. water

<h2><em>H</em><em>ope it will help you</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em></h2>
7 0
2 years ago
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