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Nezavi [6.7K]
2 years ago
13

There are many different types of models. Look at the three models. How are the models alike? How are they different?

Chemistry
2 answers:
cluponka [151]2 years ago
7 0

Answer:

Models 2 and 3 mind giving me branliest again?

Explanation:

charle [14.2K]2 years ago
5 0

Answer:

Models 2 and 3

Explanation:

Just makes sense. Hope this helps.

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If you need to measure the volume of liquid in a bottle of eyedrops, what unit would be the most practical?
nexus9112 [7]
The most practical would be milliliters
4 0
2 years ago
Read 2 more answers
Write the symbol for every chemical element that has atomic number greater than 55 and atomic mass less than 144.0 u
laila [671]
You will need a periodic table to help you answer this problem. The atomic numbers are arrange from lowest to highest in the periodic table. You can locate element number 55 to be Cesium with an atomic weight of 132.905 amu. So, you start from element 56. The following elements are:

56     Barium       137.328 amu
57     Lanthanium   138.905 amu
58     Cerium       140.116 amu
59     <span>Praseodymium    140.908 amu
60     Neodymium   144.243 amu

Neodymium is already greater than 144 amu. Therefore, these elements only include Barium, Lanthanium, Cerium and Praseodymium.</span>
3 0
3 years ago
Calcium carbide ( CaC2) reacts with water to produce acetylene (C2H2) as shown in the unbalanced reaction below: CaC2(s)+H2O(g)-
MaRussiya [10]
Answer:
46.3g H2O

Explanation:
start by balancing it: CaC2(s) + 2H2O(g) -> Ca(OH)2(s) + C2H2(g)

then use factor label method to solve

82.4g CaC2 x (1 mol CaC2/64.10g CaC2) x (2 mol H2O/1 mol CaC2) x (18.016g H2O/1 mol H20) = 46.3g H2O
4 0
3 years ago
Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2
PIT_PIT [208]

<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]

We are given:

\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ

Hence, the standard heat for the given reaction is -138.82 kJ

3 0
3 years ago
How does the arrangement of elements in periods relate to electron configuration
VashaNatasha [74]
Electronic Configuration of elements in a period is same because If you see the electronic Configuration of elements in a period you will notice that the valence shell electrons for all elements are present in the same Shell. For example, in first period consisting of Hydrogen and Helium, both the elements' valence electrons are present in the same Shell.
Electronic Configuration of Hydrogen,
1s^1
Electronic Configuration of Helium,
1s^2

Both elements' valance electrons are present in the 1st shell

(This is just a small example to understand the concept because other periods are long but the first period is short that's why I gave the example of the first period)
4 0
3 years ago
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