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LenaWriter [7]
3 years ago
5

How many moles are in 20 grams of sodium

Chemistry
1 answer:
tatyana61 [14]3 years ago
7 0
0.869952156982 to be precise
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What is the pH when the poH value is 2?
dimaraw [331]

Answer:

i hate my life  PERIODT

Explanation:

WHY AM I HERE :/

6 0
3 years ago
When 20 grams of KCIO3, is dissolved in 100 grams of water, the solution can be
tia_tia [17]

Answer:

b. unsaturated .

Explanation:

Hello there!

In this case, according to the given information, it turns out necessary for us to bear to mind the definition of each type of solution:

- Supersaturated solution: comprises a large amount of solute at a temperature at which it will be able to crystalize upon standing.

- Unsaturated solution: is a solution in which a solvent is able to dissolve any more solute at a given temperature.

- Saturated solution can be defined as a solution in which a solvent is not capable of dissolving any more solute at a given temperature.

In such a way, since 20 grams of the solute are less than the solubility, we infer this is b. unsaturated, as 33.3 grams of solute can be further added to the 100 grams of water.

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6 0
2 years ago
One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate
Ugo [173]

Answer:

\delta H_{rxn} = -66.0  \ kJ/mole

Explanation:

Given that:

3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \  \ \delta H = -47.0 \ kJ/mole  -- equation (1)  \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)}  \ \ \delta H = -25.0 \ kJ/mole  -- equation (2)  \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole  -- equation (3)

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

3FeO_{(s)} + CO_{2(g)}    \to    Fe_3O_4_{(s)} + CO_{(g)}   \ \delta H = -19.0 \ kJ/mole  -- equation (4)

Multiplying  (2) with equation (4) ; we have:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

From equation (1) ; multiplying (-1) with equation (1); we have:

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

From equation (2); multiplying (3) with equation (2); we have:

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

Now; Adding up equation (5), (6) & (7) ; we get:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

<u>                                                                                                                      </u>

FeO  \ \ \ +  \ \ \ CO   \ \  \to   \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \  \delta H = - 66.0 \ kJ/mole

<u>                                                                                                                     </u>

<u />

\delta H_{rxn} = \delta H_1 +  \delta H_2 +  \delta H_3    (According to Hess Law)

\delta H_{rxn} = (-38.0 +  47.0 + (-75.0)) \ kJ/mole

\delta H_{rxn} = -66.0  \ kJ/mole

8 0
3 years ago
Which tool is most suitable for a chemist to measure the mass of a liquid
I am Lyosha [343]

Chemists use beakers, flasks, burets and pipets to measure the volume of liquids. Plz mark as brainliest

7 0
3 years ago
Read 2 more answers
If 21.5 mol of an ideal gas is at 9.65 atm at 10.90 ∘C, what is the volume of the gas?
mamaluj [8]
<h3>Answer:</h3>

51.93 L

<h3>Explanation:</h3>

From the question we are given the following components of an ideal gas;

Number of moles = 21.5 mol

Pressure, P = 9.65 atm

Temperature, T = 10.90°C, but K= °C + 273.15

                         =284.05 k

We are required to calculate the volume of the ideal gas.

We are going to use the ideal gas equation which is given by;

PV = nRT, where P, V, T and n are the pressure, volume, temperature and moles of the ideal gas respectively. R is the ideal gas constant, 0.082057 L.atm/mol.K

To get the volume, we rearrange the formula to get;

V = nRT ÷ P

  = (21.5 × 0.082057 × 284.05 K) ÷ 9.65 atm

  = 51.93 L

Thus, the volume of the ideal gas is 51.93 L

3 0
3 years ago
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