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pogonyaev
3 years ago
8

How many grams are in 5.32 x 10^22 molecules of CO2?

Chemistry
1 answer:
ollegr [7]3 years ago
5 0

Answer:

mass = 3.89 g

Explanation:

hope this helps

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SEP Analyze Data Use graphing software or draw a graph from the data in the table. Label "Heat input (joules)" on the y-axis and
galben [10]

The characteristics of the graphic representation are to find the scales and make the best graphic, in the attachment we have the points plotted

The graphical representation is one of the best methods to see the relationships between groups of data and to be able to find the functional relationships between them.

In this case, it is requested to make a scatter plot where the measured temperature is placed on the x-axis and the heat on the y-axis.

In the attachment we can see a graph with the requested data, the most important part of finding these graphs is looking for the scales

x-axis

          scale x = \frac{\Delta Values}{paper length}

y-axis

          scale y = \frac{\Delta Values}{paper \ lengthy}

In general, the paper is 20 x 30 cm, we select the shortest part as the x axis

         scale x = \frac{50-30}{20}

         scale x = 1º  \frac{ C}{cm \ peper}  

         

Let's find the scale on the axis and the maximum value is 54000 and the minimum value is 3800, as the minimum value is much less than the maximum value, let's set this value to zero

         scale y = \frac{54000-0}{30}

         scale y = 1800  \frac{j}{cm \ paper}

To facilitate marking the values, we select a slightly larger scale,

Selected scale

           scale y = 2000 \frac{j}{cm \ paper}

Let's mark the points.

From observing these graph we see;

  • The relationships are linear
  • For the same material the amount of heat necessary to reach the same temperature is proportional to the mass of the material

Using the characteristics of the graphical representation, it is possible to find the scales and make the best graph

Learn more here: brainly.com/question/12511806

6 0
3 years ago
When of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing poi
Nesterboy [21]

The given question is incomplete. The complete question is:

When 282. g of glycine (C2H5NO2) are dissolved in 950. g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 282. g of ammonium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for ammonium chloride in X.

Answer:  the van't Hoff factor for ammonium chloride is 1.74

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point  

K_f = freezing point constant = ?

i = 1 ( for non electrolyte)

m= molality

8.2^0C=1\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of solute (glycine) = 75.07 g/mol

Mass of solute (glycine) = 282 g

8.2^0C=1\times K_f\times \frac{282g}{75.07g/mol\times 0.95kg}

K_f=2.07

ii) 20.0^0C=i\times \times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of solute (ammonium chloride) = 53.49 g/mol

Mass of solute (ammonium chloride) = 282 g

20.0^0C=i\times 2.07\times \frac{282g}{53.49g/mol\times 0.95kg}

i=1.74

Thus the van't Hoff factor for ammonium chloride is 1.74

4 0
3 years ago
In the following reaction, what coefficient will be written before the oxygen molecule (O2) to balance the chemical equation? (1
Sergeu [11.5K]
Should be C5H12 + 8O2 yields 5CO2 + 6H2O
3 0
3 years ago
Read 2 more answers
For the decomposition of ammonia on a platinum surface at 856 °C
Crazy boy [7]

\\ \tt\leadsto \dfrac{d[NH_3]}{dt}=1.50\times 10^{-6}

  • dt remains same for reaction

\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}\dfrac{d[NH_3]}{dt}

\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}(1.5\times 10^{-6})

\\ \tt\leadsto \dfrac{d[H_2]}{dt}=2.25\times 10^{-6}Ms^{-1}

M is molarity here not metre

5 0
2 years ago
A process that absorbs heat is a(n) _____ process.
kifflom [539]
B) Endothermic - It takes in heat and the temperature decreases, therefore becomes cooler.
8 0
4 years ago
Read 2 more answers
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