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stiv31 [10]
4 years ago
9

Which element has the element configuration 1s2 2s2 2p4

Chemistry
1 answer:
expeople1 [14]4 years ago
3 0

Answer:

O(Oxygen)

Explanation:

2+2+4=8

And the eighth element of the periodic table is Oxygen

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Predict the major product of the following reaction of 1-methylcyclohexene with dilute HCl in H2O?
Scrat [10]

Answer:

1-chloro-1-methylcyclohexane

Explanation:

The hydrocarbons with one double bond are called alkenes and are named with the suffix "ene". The alkenes, such as 1-methylcyclohexene, react better in an addition reaction. The double bond will be broken, and the substitutes will be placed at them.

In the reaction with HCl, H and Cl will be added to the carbons of the broken bond. The major product is formed when H is put at the carbon less substituted. Thus, the product will be the one given below, 1-chloro-1-methylcyclohexane.

7 0
3 years ago
Which has more thermal energy, an ice berg or a hot cup of coffee?
blagie [28]

Answer: an iceberg

Explanation: the reason for this is because it has much more mass then a hot cup of coffee even the the temputure of coffee is much warmer

8 0
3 years ago
What is the best way to<br> determine which brand of<br> paper towel is the<br> strongest when wet?
luda_lava [24]

Answer:Bounty

Explanation:After being soaked, the Bounty towel held an impressive 43 ounces or 2.69 pounds.

Did this help?

4 0
3 years ago
For every 5.00 mL of milk of magnesia there are 400. mg of magnesium hydroxide. How many mL of milk of magnesia do we need to ne
trasher [3.6K]
Mg(OH)2 + 2 HCl → MgCl2 + 2 H2O 

(0.0500 L) x (0.500 mol/L HCl) x (1 mol Mg(OH)2 / 2 mol HCl) x (58.3197 g Mg(OH)2/mol) = 0.729 g = 729 mg Mg(OH)2 

<span>(729 mg Mg(OH)2) x (5.00 mL / 400 mg) = 9.11 mL milk of magnesia</span>
7 0
3 years ago
Read 2 more answers
What would be the volume in liters of an 2.45 liter sample of gas at 594 °C
Ber [7]

Answer:

V_2= 1.19L

Explanation:

Hello there!

In this case, since the STP conditions are defined by 1 atm (101.3 kPa) and 273 K, it is possible for us to use the combined gas law for this problem as we are given variable pressure, temperature and volume:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

In such a way, solving for V2 as the final volume, we obtain:

V_2=\frac{P_1V_1T_2}{P_2T_1} =\frac{156kPa*2.45L*273.15K}{101.3kPa*(594+273)K}\\\\V_2= 1.19L

Regards!

6 0
3 years ago
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