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jok3333 [9.3K]
3 years ago
15

When two or more substances combine, but each keeps its own properties, the new combination is called a(an) A) Element B) Mixtur

e C) Compound D) Homogeneous substance
Chemistry
2 answers:
soldier1979 [14.2K]3 years ago
7 0
It is a mixture because it is reversible 
liraira [26]3 years ago
3 0
They form a Compound element
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When matter is broken down physically by wind or chemically by salt water, what is this known as?
Allushta [10]
C, erosion. Erosion is the natural process of breaking down natural products through wind, or similar natural resources. 
3 0
3 years ago
What is the molecular formula of a compound containing 89% cesium (Cs) and 11% oxygen (O) with a molar mass = 298 g/mol?
Maurinko [17]
M(Cs)=133 g/mol
M(O)=16 g/mol
M(CsxOy)=298 g/mol
w(Cs)=0.89
w(O)=0.11

CsxOy

x=M(CsxOy)w(Cs)/M(Cs)
x=298*0.89/133=2

y=M(CsxOy)w(O)/M(O)
y=298*0.11/16=2

Cs₂O₂  cesium peroxide
6 0
3 years ago
Read 2 more answers
Which of the following electron configurations gives the correct arrangement of the four valence electrons of the carbon atom in
Juliette [100K]

Answer:

D) 2s^12p^3

Explanation:

Carbon.

The electronic configuration is -  

1s^22s^22p^2

Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, Carbon has 2 singly occupied orbitals.

But in methane, CH_4 it forms 4 bonds. So, 1 electron each from 2s orbital jumps to the next orbital in the p subshell.

Thus, the configuration is:-

1s^22s^22p^2

Thus, the valence electron configuration is:-

2s^12p^3

7 0
2 years ago
In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what
melomori [17]

Answer:

Y=65.7\%

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2

Thus, we compute the percent yield as:

Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%

Best regards.

8 0
3 years ago
If you made 6 moles of NO2 How many grams of N2 did you use N2+2O2> 2NO2​
denis23 [38]

Answer: 1:2

Explanation:

Believe me its correct.

7 0
2 years ago
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