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2 years ago

Most elements occur in nature as a single isotope. TRUE FALSE

Chemistry

2 answers:

cestrela7 [59]2 years ago

8 0

Answer:

false

Explanation:

because elements never exist in nature as single isotopes they are always combined

lubasha [3.4K]2 years ago

4 0

Answer:

FALSE

Explanation:

in the cases of three elements (tellurium, indium, and rhenium) the most abundant isotope found in nature is actually one (or two) extremely long-lived radioisotope(s) of the element, despite these elements having one or more stable isotopes.

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GivenL C2H2 + 5O2 ------> 4 CO2 + 2H2O

diamong [38]

Answer:

101 L

Explanation:

35.0 g KOH ÷ 56.09 g/mol KOH × (1 mol H2O/ 1 mol KOH) × 18 g/mol H2O = 11.2 g H2O

35.0 g HCl ÷ 36.45 g/mol HCl × (1 mol H2O/ 1 mol HCl) × 18 g/mol H2O = 17.3 g H2O

35.0 g KOH is the limiting reactant

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3 years ago

What Is the answers?

Yuki888 [10]

Answer:

oxygen is responsible for rusting

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2 years ago

Read 2 more answers

Identify the reactants and the products of the preparation step to the urea cycle.

Salsk061 [2.6K]

According to sources, the most probable answer to this query is the enzymes and waste products that are collected by the nephron from the blood. Thank you for your question. Please don't hesitate to ask in Brainly your queries.

7 0

2 years ago

This is for a study guide, I can't figure it out!

miskamm [114]

<span>The answer is "D" where the number of collisions per unit area is reduced by one-half. Drawing back on the piston means the volume is increased. The pressure is reduced. There are fewer collisions when the pressure is reduced.</span>

3 0

3 years ago

Write the balanced standard combustion reaction for the c5h7on3s2

AlexFokin [52]

Answer:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Explanation:

When C5H7ON3S2 under go Combustion, the following are obtained as illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

Now, let us balance the equation. This is illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

There are 3 atoms of N on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of C5H7ON3S2 and 6 in front of N2 as shown below:

4C5H7ON3S2 + O2 —> CO2 + H2O + 6N2 + S2

There are 20 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 20 in front of CO2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + H2O + 6N2 + S2

There are 28 atoms on H on the left side and 2 atoms on the right side. It can be balance by putting 14 in front of H2O as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + S2

There are 8 atoms of S on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of S2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, there are a total of 54 atoms of O on the right side and 6 atoms on the left side

It can be balance by putting 25 in front of O2 as shown below:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, we can see that the equation is balanced.

4 0

2 years ago