Are soft drink, solder and air homogeneous or heterogeneous?

Coal is transported from mine to plant via “unit trains” that usually consist of 100 cars each carrying 100 tons of coal. If one

abruzzese [7]

I think it’s mass number 12

3 0

3 years ago

What are the controlled variables for water of crystallization experiment?

yarga [219]

Given what we know, we can confirm that as with any experiment, the control variable will be the one that through each trial of the experiment, no matter how many times it is performed, stays constant.

<h3>What is a controlled variable?</h3>

A variable that remains constant through an experiment.
They are used to compare results to the normal condition.
They are also used to isolate the changes to one factor at a time and thus know its exact effects on the outcome.
This increases the accuracy of the data and the subsequent conclusion.

Therefore, we can confirm that if a variable stays constant through each phase and trial of an experiment, it is considered to be a controlled variable and is useful in order to increase the accuracy of the conclusion.

To learn more about control variables visit:

brainly.com/question/17555102?referrer=searchResults

7 0

3 years ago

Balance the following equations:

lesya [120]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2>

(I) $N_{2} + 3H_{2} --> 2NH_{3}$

(II) $P_{4} + 5O_{2} --> 2P_{2} O_{5}$

(III) $2NaF + Br_{2} -->2NaBr + F_{2}$

(IV) $2ZnS + 3O_{2} --> 2ZnO + 2SO_2$

(V) $Pb(NO_3)_2 + 2NaCl --> 2NaNO_3 + PbCl_2$

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

7 0

3 years ago

Determine the mass of CO2 gas produced when 8.24 g of NaHCO3 is added to a solution that contains 4.41 g of HCl.

hjlf

The mass of CO₂ gas = 4.312 g

<h3>Further explanation</h3>

Reaction

NaHCO₃ (s) + HCl (aq) → NaCl (aq) + H₂O (aq) + CO₂ (g)

Find limiting reactant

mol NaHCO₃

$\tt \dfrac{8.24}{84,007 }=0.098$

mol HCl

$\tt \dfrac{4.41}{36.5}=0.121$

Limiting reactant : NaHCO₃ (smaller)

mol CO₂ = mol NaHCO₃ = 0.098

mass CO₂ :

$\tt 0.098\times 44=4.312~g$

4 0

3 years ago

Unit: Stoichiometry

Reika [66]

Answer:

1. 2.41 × 1023 formula units

2. 122 L

3. 7.81 L

Explanation:

1. Equation of the reaction: 2 Na(NO3) + Ca(CO3) ---> Na2(CO3) + Ca(NO3)2

Mole ratio of NaNO3 to CaCO3 = 2 : 1

Moles of CaCO3 = mass/molar mass

Mass of CaCO3 = 20 g; molar mass of CaCO3 = 100 g

Moles of CaCO3 = 20 g/100 g/mol = 0.2 moles

Moles of NaNO3 = 2 × 0.2 moles = 0.4 moles

1 Mole of NaNO3 = 6.02 × 10²³ formula units

0.4 moles of NaNO3 = 0.4 × 6.02 × 10²³ = 2.41 × 1023 formula units

2. Equation of reaction : 2 H2O ----> 2 H2 + O2

Mole ratio of oxygen to water = 1 : 2

At STP contains 6.02 × 10²³ molecules = 1 mole of water

6.58 × 10²⁴ molecules = 6.58 × 10²⁴ molecules × 1 mole of water/ 6.02 × 10²³ molecules = 10.93 moles of water

Moles of oxygen gas produced = 10.93÷2 = 5.465 moles of oxygen gas

At STP, 1 mole of oxygen gas = 22.4 L

5.465 moles of oxygen gas = 5.465 moles × 22.4 L/1 mole = 122 L

3.Equation of reaction: 6 K + N2 ----> 2 K3N

Mole ratio of Nitrogen gas and potassium = 6 : 1

Moles potassium = mass/ molar mass

Mass of potassium = 90.0 g, molar mass of potassium = 39.0 g/mol

Moles of potassium = 90.0 g / 39.0 g/mol = 2.3077moles

Moles of Nitrogen gas = 2.3077 moles / 6 = 0.3846 moles

At STP, 1 mole of nitrogen gas = 22.4 L

0.3486 moles of oxygen gas = 0.3486 moles × 22.4 L/1 mole = 7.81 L

7 0

3 years ago