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3 years ago

The usefulness of mendeleev s periodic table was confirmed by

1 answer:

7 0

In the table of mendeleev, the lines feature the elements in increasing order of atomic mass.

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A 0.08024 M solution of NaOH was used to titrate a solution of unknown concentration of HCl. A 32.08 mL sample of the HCl soluti

iris [78.8K]

Answer:

0.08024

Explanation:

4 0

3 years ago

Element X reacts with chlorine to form an ionic compound that has the formula XCl₂. To which group on the Periodic Table could e

marishachu [46]

Answer:

Group 2 is the correct answer.

Explanation:

Chlorine has a charge -1. It means that X is +2 so that the formula is XCl2.

4 0

3 years ago

Which element consists of positive ions immersed in a “sea” of mobile electrons

worty [1.4K]

Answer:

Explanation:

Nickel is the only metal listed. Metal elements have metallic bonds between the metal atoms. Positive metal ions in a sea of electrons.

4 0

3 years ago

Oxalic Acid, a compound found in plants and vegetables such as rhubarb, has a mass percent composition of 26.7% C, 2.24% H, and

blondinia [14]

Answer:

HCO₂

Explanation:

From the information given:

The mass of the elements are:

Carbon C = 26.7 g; Hydrogen H = 2.24 g Oxygen O = 71.1 g

To determine the empirical formula;

First thing is to find the numbers of moles of each atom.

For Carbon:

$=26.7 \ g\times \dfrac{1 \ mol }{12.01 \ g} \\ \\ =2.22 \ mol \ of \ Carbon$

For Hydrogen:

$=2.24 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =2.22 \ mol \ of \ Hydrogen$

For Oxygen:

$=71.1 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =4.44 \ mol \ of \ oxygen$

Now; we use the smallest no of moles to divide the respective moles from above.

For carbon:

$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ carbon$

For Hydrogen:

$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ hydrogen$

For Oxygen:

$\dfrac{4.44 \ mol \ of \ Oxygen}{2.22} =2 \ mol \ of \ oxygen$

Thus, the empirical formula is HCO₂

4 0

3 years ago

A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4

Harman [31]

Na₃PO₄ -----> 3Na(+) + PO₄(3-)
y-y.......................3y...........y

3y = 1.2
y = 0,4M

Na₃PO₄ -----> 3Na(+) + PO₄(3-)
0,4-0,4..............1,2..........0,4
0.........................1,2..........0,4

C = n/V
n = C×V
n = 0,4×0,65L
n = 0,26 mol Na₃PO₄

mNa₃PO₄: (23×3)+31+(16×4) = 164 g/mol

164g ----- 1 mol
Xg -------- 0,26 mol
X = 164×0,26
X = 42,64g Na₂SO₄

5 0

3 years ago