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3 years ago

What’s the answer? And if you find the answer in a pdf, comment the link.

Chemistry

1 answer:

dimulka [17.4K]3 years ago

4 0

You can simply look at the order of the elements on a periodic table. Therefore, it is B)

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What is the effect of decreasing concentration of sodium hydroxide ten times in standardization

tatuchka [14]

Answer:

[Translation] pH will decrease by 1.

Explanation:

See attached image. The translation is in the "Answer".

7 0

3 years ago

Iridium has only two naturally occurring isotopes. The mass of iridium-191 is 190.9605 amu and the mass of iridium-193 is 192.96

cluponka [151]

Answer:

Abundance of Iridium-193 is 62.75%

Explanation:

From the question given above, the following data were obtained:

Isotope A (Iridium-191):

Mass of A = 190.9605 amu

Abundance of A = A%

Isotope B (Iridium-193):

Mass of B = 192.9629 amu

Abundance B = (100 – A) %

Relative atomic mass of Iridium = 192.217 amu

Next, we shall determine the abundance of isotope A (Iridium-191). This can be obtained as follow:

Relative atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]

192.217 = [(190.9605 × A%)/100] + [(192.9629 × (100 – A)%)/100]

192.217 = 1.909605A% + 1.929629(100 – A)%

192.217 = 1.909605A% + 192.9629 – 1.929629A%

Collect like terms

192.217 – 192.9629 = 1.909605A% – 1.929629A%

–0.7459 = –0.020024A%

Divide both side by –0.020024

A% = –0.7459 / –0.020024

A% = 37.25 %

Finally, we shall determine the abundance of Isotope B (Iridium-193).

This can be obtained as follow:

Abundance of A (Iridium-191) = 37.25 %

Abundance of B (Iridium-193) =?

Abundance B = 100 – A%

Abundance B = 100 – 37.25 %

Abundance of B (Iridium-193) = 62.75%

4 0

2 years ago

A student prepares a 100.0 mL solution using 44.7 grams of potassium nitrite. They then take 11.9 mL of this solution and dilute

Naddik [55]

Answer:

0.52 g of KNO₃ are contained in 19.7 mL of diluted solution.

Explanation:

We can work on this problem in Molarity cause it is more easy.

Molarity (mol/L) → moles of solute in 1L of solution.

100 mL of solution = 0.1 L

We determine moles of solute: 44.7 g . 1mol /101.1 g = 0.442 mol of KNO₃

Our main solution is 0.442 mol /0.1L = 4.42 M

We dilute: 4.42 M . (11.9mL / 200mL) = 0.263 M

That's concentration for the diluted solution.

M can be also read as mmol/mmL, so let's find out the mmoles

0.263 M . 19.7mL = 5.18 mmol

We convert the mmol to mg → 5.18 mmol . 101.1 mg / mmol = 523.7 mg

Let's convert mg to g → 523.7 mg . 1 g / 1000 mg = 0.52 g

6 0

3 years ago

How do you label the delta E, on an energy diagram

vladimir1956 [14]

Delta energy on labelled diagram is attached below

4 0

3 years ago

Consider a bent molecule, such as H2Se, in which the central atom has two lone pairs of electrons. The electronegativities of H

Shalnov [3]

Answer:

Based on these values and on consideration of molecular geometry, the H-Se bond can be considered almost _____non-polar___ and the molecule is __polar_____.

Explanation:

Looking at the difference in electro negativity of the two elements; hydrogen and selenium, one may be led to the conclusion that the molecule is nonpolar since the magnitude of electronegative between the two bonding atoms is minimal.

However, electro negativity difference alone is insufficient to determine the polarity of a molecule. The structure of the molecule is also considered. Based on the structure of the molecule, it is expected to have a dipole moment. Hence the molecule is polar.

7 0

3 years ago