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podryga [215]
2 years ago
10

What do atom, molecules, and elements have in common

Chemistry
1 answer:
____ [38]2 years ago
6 0

Answer:

Solids, liquids, gases - all matter - are made up of atoms (or other things, like molecules, that are made from atoms)! ELEMENTS are the kinds of atoms that we can have. Carbon is an element, hydrogen is an element, and so is oxygen. ... Atoms can join together - they form bonds together - to make MOLECULES.

You might be interested in
How many moles of HNO3 will be produced 3 NO2+H2O=2HNO3+ NO
valkas [14]

Answer:

2 moles of HNO3

Explanation:

The equation seems to be balanced correctly. The problem is we done know what you started with. We will assume it is 3 moles of NO2.

If that is the case then 2 moles of HNO3 will be produced.

8 0
3 years ago
If 25.16 g of chlorine react with 12.99 g of manganese metal, what is the empirical formula of the compound?
WARRIOR [948]

We use the given masses of the reactants to calculate the moles of Mn and Cl. Empirical formula represents the simplest mole ratio of atoms present in a compound.

Moles of Mn = 12.99 g Mn * \frac{1 mol Mn}{54.94 g Mn} = 0.236 mol Mn

Moles of Cl = 25.16 g Cl_{2} *\frac{1 mol Cl_{2}}{70.91 g Cl_{2}} * \frac{2 mol Cl}{1 mol Cl_{2}} = 0.710 mol Cl

Simplest mole ratio: Mn_{\frac{0.236}{0.236}}Cl_{\frac{0.710}{0.236}}

So the empirical formula is MnCl_{3}

6 0
3 years ago
What is the melting point of a solution in which 3.5 grams of sodium chloride is added to 230 mL of water?
kirza4 [7]
We are going to use this equation:

ΔT = - i m Kf

when m is the molality of a solution 

i = 2

and ΔT is the change in melting point = T2- 0 °C

and Kf is cryoscopic constant = 1.86C/m

now we need to calculate the molality so we have to get the moles of NaCl first:

moles of NaCl = mass / molar mass

                         = 3.5 g / 58.44 

                         = 0.0599 moles


when the density of water = 1 g / mL and the volume =230 L

∴ the mass of water = 1 g * 230 mL = 230 g = 0.23Kg 

now we can get the molality = moles NaCl / Kg water

                                                =0.0599moles/0.23Kg

                                                = 0.26 m

∴T2-0 =  - 2 * 0.26 *1.86

∴T2 = -0.967 °C
8 0
3 years ago
Read 2 more answers
At 25C the density of water is 0.997044 g/mL. Use this value to determine the percent error for the two density measurements
Gnom [1K]

Given that:

  • At 25C the density of water is 0.997044 g/mL.

From the information attached below, we have the following parameters.

The density of water calculation using a bottle.

     Initial volume of    Final volume of    Mass of water   Density (g/mL)

     burette (mL)        burette   (mL)       dispensed (g)

 

Sample 1      2.33                     7.34                   5.000               -----

Sample 2      7.34                    12.37                 5.025                -----

Sample 3      12.37                   18.50                6.112                  -----

Sample 4      18.50                  24.57               6.064                 -----

Sample 5     24.57                  31.31                6.720                  -----

The first thing we need to do is to determine the change in the volume of the burette in each sample from the above information.

  • The change in the volume of the burette = (final volume - the initial volume) mL

Sample 1:

= (7.34 - 2.33) mL

= 5.01 mL

Sample 2:

= (12.37 - 7.34) mL

= 5.03 mL

Sample 3:

= (18.50 - 12.37) mL

= 6.03 mL

Sample 4:

= (24.57 - 18.50) mL

= 6.07 mL

Sample 5:

= (31.31 - 24.57) mL

= 6.74 mL

The mass of the water dispersed in sample 1 is given as = 5.000 g

Using the relation for calculating the density of each, we have:

Sample 1

\mathbf{density = \dfrac{mass}{volume}}

\mathbf{density = \dfrac{5.01 g}{5.000 ml}}

density = 0.998004 g/ml

Sample 2:

\mathbf{density = \dfrac{5.025 g}{5.03ml}}

density = 0.999006 g/ml

Sample 3:

\mathbf{density = \dfrac{6.112 g}{6.13ml}}

density = 0.997064 g/ml

Sample 4:

\mathbf{density = \dfrac{6.064 \ g}{6.07 \ ml}}

density = 0.999012 g/ml

Sample 5:

\mathbf{density = \dfrac{6.720 \ g}{6.74 \ ml}}

density = 0.997033 g/ml

Thus, the average density for all the samples is:

\mathbf{= \dfrac{( 0.998004 + 0.999006 + 0.997064 +   0.999012  + 0.997033  )}{5}}

= 0.998024

∴

The percentage error for the two densities measurement is:

=\dfrac{ (experimental \  value -theoretical  \ value)\times 100 }{theoretical  \ value}

Given that the theoretical value = 0.997044 g/ml

Then;

\mathbf{= \dfrac{(0.998024 - 0.997044)100}{0.997044}}

= 0.0983%

Therefore, we can conclude that the percent error for the two density measurements is 0.0983%

Learn more about density here:

brainly.com/question/24386693?referrer=searchResults

4 0
2 years ago
3 waves are shown with a line through their center. The bottom of the first wave is labeled C. A bracket labeled D connects the
Likurg_2 [28]

Answer:Label the parts of this wave.

A:  

✔ crest

B:  

✔ amplitude

C:  

✔ trough

D:  

✔ wavelength

Explanation:

8 0
2 years ago
Read 2 more answers
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