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3 years ago

What particles in an atom are involved in nuclear reactions

Chemistry

1 answer:

VikaD [51]3 years ago

8 0

Answer:

Explanation:

Electrons are the negatively charged particles of an atom that are involved in chemical reactions. They are located outside the nucleus of an atom

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Acetylene gas (c2h2) reacts with oxygen gas (o2) to produce carbon dioxide (co2) and water vapor (h2o). how many liters of c2h2

MissTica

According to the balanced equation of the reaction:

2C2H2 + 5O2 → 4CO2 + 2H2O

So we can mention all as liters,

A) as we see that 2 liters of C2H2 react with 5 liters of oxygen to produce 4 liters of CO4 and 2 liters of H2O

So, when we have 75L of CO2

and when we have 2 L of C2H2 reacts and gives 4 L of CO2

2C2H2 → 4CO2

∴ The volume of C2H2 required is:

= 75L / 2

= 37.5 L

B) and, when we have 75 L of CO2

and 4CO2 → 2H2O

∴ the volume of H2O required is:

= 75 L /2
= 37.5 L

C) and from the balanced equation and by the same way:

when 5 liters O2 reacts to give 4 liters of CO2

and we have 75 L of CO2:

5 O2 → 4 CO2

?? ← 75 L

∴ the volume of O2 required is:

= 75 *(5/4)

= 93.75 L

D) about the using of the number of moles the answer is:

no, there is no need to find the number of moles as we called everything in the balanced equation by liters and use it as a liter unit to get the volume, without the need to get the number of moles.

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3 years ago

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1,670,000,000 in<br> scientific notation pleaseeee

noname [10]

Answer:

1.67 × 10^9

Explanation:

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2 years ago

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10. When a 13.6 g sample of a compound containing only magnesium and oxygen is decomposed, 5.4 g of oxygen is obtained. What is

irinina [24]

Answer:5.4 g / 13.6 g *100

Explanation:Its is the correct answer

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3 years ago

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A student is shown a simple model of particles that make up a liquid, as shown

elena55 [62]

E⁣⁣⁣xplanation i⁣⁣⁣s i⁣⁣⁣n a f⁣⁣⁣ile

bit. $^{}$ ly/3a8Nt8n

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2 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago