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3 years ago

9

In details explain element

2 answers:

Mamont248 [21]3 years ago

7 0

Going off of what the other person said
An element can also be “each of more than one hundred substances that cannot be chemically interconverted or broken down into simpler substances and are primary constituents of matter. Each element is distinguished by its atomic number, i.e. the number of protons in the nuclei of its atoms.”

Dominik [7]3 years ago

6 0

Answer:

definition- a part or aspect of something abstract, especially one that is essential or characteristic

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What mechanical weathering process is used when water turns to ice?

Black_prince [1.1K]

ANSWER: ice wedging, also known as freeze-thaw weathering

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3 years ago

Read 2 more answers

You add 168.90 grams of NaCl to a container and then you add 616.00 grams of water to that same container What is the weight per

erma4kov [3.2K]

Answer:

The weight percent of NaCl in the container is 21.5%

Explanation:

Given that,

Mass of NaCl = 168.90 gram

Mass of water = 616.00 grams

We need to calculate the weight percent of NaCl in the container

Using formula of percentage of weight

$weight\ \%\ of\ component\ of\ the\ solution =\dfrac{weight\ of\ the\ component\ in\ the\ solution}{total\ weight\ of\ the\ solution}\times100$

Put the value into the formula

$weight\ \%\ of\ component\ of\ the\ solution =\dfrac{168.90}{168.90+616.00}\times100$

$weight\ \%\ of\ component\ of\ the\ solution=21.5\%$

Hence, The weight percent of NaCl in the container is 21.5%

5 0

3 years ago

4 Al + 3O2 → 2Al2O3 If 14.6 grams Al are reacted, how many liters of O2 at STP would be required?

melomori [17]

Answer: 9.08 L

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}$

$\text{Moles of} Al=\frac{14.6g}{27g/mol}=0.54moles$

$4Al+3O_2\rightarrow 2Al_2O_3$

According to stoichiometry :

4 moles of $Al$ require = 3 moles of $O_2$

Thus 0.54 moles of $Al$ will require= $\frac{3}{4}\times 0.54=0.405moles$ of $O_2$

Standard condition of temperature (STP) is 273 K and atmospheric pressure is 1 atm respectively.

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 1 atm

V= Volume of the gas = ?

T= Temperature of the gas = 273 K

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas= 0.405

$V=\frac{nRT}{P}=\frac{0.405\times 0.0821\times 273}{1}=9.08L$

Thus 9.08 L of $O_2$ at STP would be required

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3 years ago

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rodikova [14]

Answer:

True

Explanation:

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Answer:

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