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kozerog [31]
3 years ago
8

What are atoms in chemistry?​

Chemistry
2 answers:
kobusy [5.1K]3 years ago
3 0

Answer:

Smallest living thing

Explanation:

Annette [7]3 years ago
3 0

Answer:

In chemistry, atom is the smallest particle of an element

Explanation:

Atoms are smallest particle of an element.

For example, when you have a substance, lets say an apply, and you keep on cutting it, It will reach a point in which you will be unable to cut it anymore, at that point we call the particle an atom.

Atoms are neutral in charge. Inside the atoms, there is a nucleus which consist of proton, neutron and electrons. The protons  are positively charged,  the electrons are negatively charged. There are equal number of protons and electrons in an atom, this make the atom to be neutral in charges.

Atoms are the building block of any materials or substance.

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The specific rate constant, k, for radioactive beryllium–11 is 0.049 s–1. What mass of a 0.500 mg sample of beryllium–11 remains
Sonja [21]

<u>Answer: </u>The mass of sample that remained is 0.127 mg

<u>Explanation:</u>

The integrated rate law equation for first-order kinetics:

k=\frac{2.303}{t}\log \frac{a}{a-x} ......(1)

Given values:

a = initial concentration of reactant = 0.500 mg

a - x = concentration of reactant left after time 't' = ?mg

t = time period = 28 s

k = rate constant = 0.049s^{-1}

Putting values in equation 1:

0.049s^{-1}=\frac{2.303}{28s}\log (\frac{0.500}{(a-x)})\\\\\log (\frac{0.500}{(a-x)})=\frac{0.049\times 28}{2.303}\\\\\frac{0.500}{a-x}=10^{0.5957}\\\\frac{0.500}{a-x}=3.94\\\\a-x=\frac{0.500}{3.942}=0.127mg

Hence, the mass of sample that remained is 0.127 mg

5 0
3 years ago
Estuaries are called "the nurseries of the sea" because many species lays eggs and develop their young in estuaries.Which two fe
Misha Larkins [42]
The answer would be b.
7 0
3 years ago
What is the molecular shape of HCN?<br> bent<br> linear<br> angular<br> trigonal pyramidal
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HCN is a linear molecule.
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Read 2 more answers
A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el
icang [17]

Answer:

The temperature of the air at this given elevation will be 53.32425°C

Explanation:

We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.

Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm  ; Final Volume = 1.8 L ; Final pressure = 0.73 atm  

We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T_2 ),

P_1V_1 / T_1 = P_2V_2 / T_2,

T_2 = P_2V_2T_1 / P_1V_1,

T_2 = 0.73 atm * 1.8 L * 298.15 K / 1 atm * 1.2 L = ( 0.73 * 1.8 * 298.15 / 1 * 1.2 ) K = 326.47425 K,

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7 0
3 years ago
Be sure to answer all parts.
MrRissso [65]

Answer: The molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

Explanation:

Molarity is the number of moles of a substance present in liter of a solution.

And, moles is the mass of a substance divided by its molar mass.

(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.

Moles = \frac{mass}{molar mass}\\= \frac{28.5 g}{46 g/mol}\\= 0.619 mol

Now, molarity of ethanol solution is as follows.

Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.619 mol}{4.50 \times 10^{2} \times 10^{-3}L}\\= 1.38 M

(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.

Moles = \frac{mass}{molar mass}\\= \frac{21.6 g}{342.3 g/mol}\\= 0.063 mol

Now, molarity of sucrose solution is as follows.

Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.063 mol}{0.067 L}  (1 mL = 0.001 L)\\= 0.94 M

(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.

Moles = \frac{mass}{molar mass}\\= \frac{6.65 g}{58.44 g/mol}\\= 0.114 mol

Now, molarity of sodium chloride solution is as follows.

Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.114 mol}{0.0962 L}\\= 1.182 M

Thus, we can conclude that the molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

4 0
4 years ago
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