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Arturiano [62]
4 years ago
12

What part of an atom determines if it will combine or break apart from other substances?

Chemistry
1 answer:
Anni [7]4 years ago
8 0

Electron is the correct answer :D

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how many ml of 6.00 m hcl solution is needed to make 25.0 ml of 3.00 m hcl solution? answer in ml with correct significant figur
evablogger [386]

The answer is 12.5 mL.

Solution:

<u>M, 6.00M          V₁ =?</u>

M2 = 3.00M,    V₂ = 25.0mL

<u>M1V1 = M2 V2/M1</u>

V1 = M2V2 M,

V1 = 25.0X3.00/6.00

∴ V1 = 12.5 mL

The initial concentration of the resulting solution is 2 molar which is the volume multiplied by the concentration. The volume of the starting solution is therefore 2,500 ml × 2.25 mol. Calculate the number of moles of HCl dissolved in a given volume of acid by reacting an acid with CaCO3.

This is essentially a double exchange reaction with the decomposition of one of the products. To determine the amount of stock solution required divide the number of moles of glucose by the molarity of the stock solution.

Learn more about HCl here:-brainly.com/question/27576431

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8 0
1 year ago
While running a column a student notices that after 43 min, approximately 30 mL of solvent has been eluted. Calculate the flow r
34kurt

Answer:

F = V / T

   =  \frac{30ml}{0.72hr}

    = 42ml/hr

Explanation: first convert time into hours,then deduce the formula for the flow rate

6 0
3 years ago
Can someone help I have to much stuff to do tonight so any help is appreciated.
Anit [1.1K]
Sorry I'm not in chemistry but do you have any other work
6 0
3 years ago
Which of the following atoms contains the same number of neutrons as an atom of Fluorine-19?
Pani-rosa [81]
In Flourine-19, you can determine that the number of Neutrons is 10.

Now, if you look at Neon-20, you subtract its mass by the number of protons, leaving you with 10.

Therefore, Neon-20 has the same number of neutrons as Flourine-19.
8 0
4 years ago
Read 2 more answers
Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant f
umka21 [38]

Answer:

1) 0.00498 mol Cu.

2) 0.00000374 mol CO₂

Explanation:

Question 1)

We want to convert 3.00 * 10²¹ copper atoms into moles. Note that 3.00 is three significant figures.

Recall that by definition, one mole of a substance has exactly 6.022 * 10²³ amount of that substance. In other words, we have the ratio:

\displaystyle \frac{1\text{ mol}}{6.022\times 10^{23} \text{ Cu}}

We are given 3.00 * 10²¹ Cu. To cancel out the Cu, we can multiply it by our above ratio with Cu in the denominator. Hence:

\displaystyle 3.00 \times 10^{21} \text{ Cu} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} \text{ Cu}}

Cancel like terms:

=\displaystyle 3\times 10^{21} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} }

Simplify:

\displaystyle  = \frac{3\text{ mol Cu}}{6.022 \times 10^{2}}

Use a calculator:

= 0.004981... \text{ mol Cu}

Since the resulting answer must have three significant figures:

= 0.00498\text{ mol Cu}

So, 3.00 * 10²¹ copper atoms is equivalent to approximately 0.00498 moles of copper.

Question 2)

We want to convert 2.25 * 10¹⁸ molecules of carbon dioxide into moles. Note that 2.25 is three significant digits.

By definition, there will be 6.022 * 10²³ carbon dioxide molecules in one mole of carbon dioxide. Hence:

\displaystyle \frac{6.022 \times 10^{23} \text{ CO$_2$}}{1\text{ mol CO$_2$}}

To cancel the carbon dioxide from 2.25 * 10¹⁸, we can multiply it by the above ratio with the carbon dioxide in the denominator. Hence:

\displaystyle 2.25\times 10^{18} \text{ CO$_2$} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23} \text{ CO$_2$}}

Cancel like terms:

\displaystyle= 2.25\times 10^{18} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23}}

Simplify:

\displaystyle = \frac{2.25 \text{ mol CO$_2$}}{6.022\times 10^5}}

Use a calculator:

=0.000003736...\text{ mol CO$_2$}

Since the resulting answer must have three significant figures:

= 0.00000374\text{ mol CO$_2$}

So, 2.25 * 10¹⁸ molecules of carbon dioxide is equivalent to approximately 0.00000374 moles of carbon dioxide.

4 0
3 years ago
Read 2 more answers
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