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ioda
3 years ago
6

What r the uses of bases...!??(follow me plz) ​

Chemistry
1 answer:
Fofino [41]3 years ago
6 0

Answer:

your mom hy hsin DDT tfx hutch knoll b.f dad tho

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Compare the chemical reactivity of chlorine and bromine. Explain your answer. Thanks in advance
shepuryov [24]

Bromine vs Chlorine | Br vs Cl

 

Halogens are group VII elements in the periodic table, and all are electronegative elements and have the capability to produce -1 anions.

Bromine

Bromine is denoted by the symbol Br. This is in the 4th period of the periodic table between chlorine and iodine halogens. Its electronic configuration is [Ar] 4s2 3d10 4p5. The atomic number of bromine is 35. Its atomic mass is 79.904. Bromine staChlorine is an element in the periodic table which is denoted by Cl.  It is a halogen (17th group) in the 3rd period of the periodic table. The atomic number of chlorine is 17; thus, it has seventeen protons and seventeen electrons. Its electron configuration is written as 1s2 2s2 2p6 3s2 3p5. Since the p sub level should have 6 electrons to obtain the Argon, noble gas electron configuration, chlorine has the ability to attract an electron. ys as a red-brown color liquid at room temperature.

8 0
3 years ago
Sugar<br> O2<br> CO2<br> H2O<br> What is the name of this energy pathway?
lys-0071 [83]

thnxx for free point? ?????????

7 0
2 years ago
THIS IS SCIENCE
lana [24]
I'm guessing D or C, remember that the noble gases cannot combine
5 0
3 years ago
3. A student carries out the clay-catalyzed dehydration of cyclohexanol starting with 10 moles of cyclohexanol and obtains 500 m
IrinaK [193]

Answer:

49.45~%

Explanation:

In this case, we have to start with the <u>chemical reaction</u>:

C_6H_1_2O~->~C_6H_1_0~+~H_2O

So, if we start with <u>10 mol of cyclohexanol</u> (C_6H_1_2O) we will obtain 10 mol of cyclohexanol (C_6H_1_0). So, we can calculate the grams of cyclohexanol if we<u> calculate the molar mass:</u>

(6*12)+(10*1)=82~g/mol

With this value we can calculate the grams:

10~mol~C_6H_1_0\frac{82~g~C_6H_1_0}{1~mol~C_6H_1_0}=820~g~C_6H_1_0

Now, we have as a product 500 mL of C_6H_1_0. If we use the <u>density value</u> (0.811 g/mL). We can calculate the grams of product:

500~mL\frac{0.811~g}{1~mL}=405.5~g

Finally, with these values we can calculate the <u>yield</u>:

%~=~\frac{405.5}{820}x100~=~49.45%%= (405.5/820)*100 = 49.45 %

See figure 1

I hope it helps!

6 0
3 years ago
A sample of gas (1.9 mol) is in a flask at 21 °c and 697 mm hg. the flask is opened and more gas is added to the flask. the new
garik1379 [7]

To solve this problem, we assume ideal gas so that we can use the formula:

PV = nRT

since the volume of the flask is constant and R is universal gas constant, so we can say:

n1 T1 / P1 = n2 T2 / P2

 

1.9 mol * (21 + 273 K) / 697 mm Hg = n2 * (26 + 273 K) / 841 mm Hg

<span>n2 = 2.25 moles</span>

8 0
3 years ago
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