How many protons are found in an atom of platinum (pt)?

PLEASE HELP ME!!! ASAP

shtirl [24]

Answer:

Theoretical yield of the reaction = 34 g

Excess reactant is hydrogen

Limiting reactant is nitrogen

Explanation:

Given there is 100 g of nitrogen and 100 g of hydrogen

Number of moles of nitrogen = 100 ÷ 28 = 3·57

Number of moles of hydrogen = 100 ÷ 2 = 50

Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation

N2 + 3H2 → 2NH3

From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed

Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen

⇒ We require 10·71 moles of hydrogen

But we have 50 moles of hydrogen

∴ Limiting reactant is nitrogen and excess reactant is hydrogen

From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia

Molecular weight of ammonia = 17 g

∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g

5 0

4 years ago

What is the name of the product isotope formed when Thorium-234 decays by beta decay?

defon

Answer:

protactinium-234,

Explanation:

This is what happens when thorium-234 releases a W- boson, which then decays to an electron and an electron antineutrino.

7 0

3 years ago

A meteorologist filled a weather balloon with 3.00L of the inert noble gas helium. The balloon's pressure was 765 torr. The ball

yanalaym [24]

Answer:

4.33 L

Explanation:

Step 1: Given data

Initial volume of the balloon (V₁): 3.00 L

Initial pressure of the balloon (P₁): 765 torr

Final volume of the balloon (V₂): ?

Final pressure of the balloon (P₂): 530 torr

Step 2: Calculate the final volume of the balloon

If we consider Helium to behave as an ideal gas, we can calculate the final volume of the balloon using Boyle's law.

$P_1 \times V_1 = P_2 \times V_2\\V_2 = \frac{P_1 \times V_1}{P_2} = \frac{765torr \times 3.00L}{530torr} = 4.33 L$

7 0

3 years ago

How many moles of aspartame are present in 1.00 mg of aspartame?

Diano4ka-milaya [45]

Answer:- There are $3.40*10^-^6$ moles.

Solution:- It is a unit conversion problem where we are asked to convert mg of aspartame to moles. Aspartame is $C_1_4H_1_8N_2O_5$ and it's molar mass is 294.31 grams per mole.

mg are converted to grams and then the grams are converted to moles as:

$1.00mg Aspartame(\frac{1g}{1000mg})(\frac{1mole}{294.31g})$

= $3.40*10^-^6$ moles of aspartame

So, there would be $3.40*10^-^6$ moles of aspartame in 1.00 mg of it.

3 0

4 years ago

Robert has pure samples of both D-ribose and D-arabinose, but he forgot to label them. He only has some nitric acid, the reagent

Sauron [17]

Answer:

Following are the responses to the given question:

Explanation:

Since HN03 is an oxidation substance D-ribose u.ith oxidized to form in rubric acid Ribose is chiral, but rubric acid is achiral because of its symmetry mirror level, Hence no infrared roster in the sample holder is observed.

Please find the attached file.

D-Arabinose, on either hand, gives optical aldaric acid with such a net optical rotation observed inside the polarimeter for diagnosis with HN03.

4 0

3 years ago