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2 years ago

15

A student does not observe a change when holding a test tube in a flame. However, a change is expected. What is the most likely

reason for this?

2 answers:

love history [14]2 years ago

5 0

Heat is energy, and that energy would eventually cause the object to undergo a phase change.

astraxan [27]2 years ago

5 0

D. The reactants were not heated long enough.

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Sati [7]

In a chemical reaction, the limiting reagent is the chemical being used up while the excess reactant is the chemical left after the reaction process.

Before calculating the limiting and excess reactant, it is important to balance the equation first by stoichiometry.

C25N3H30Cl + NaOH = C25N3H30OH + NaCl

Since the reaction is already balanced, we can now identify which is the limiting and excess reagent.

First, we need to determine the number of moles of each chemical in the equation. This is crucial for determining the limiting and excess reagent.

<span>Assuming that there is the same amount of solution X for each reactant</span>

1.0 M NaOH ( X ) = 1.0 moles NaOH

1.00 x 10-5 M C25N3H30Cl ( X ) = 1.00 x 10-5 moles C25N3H30Cl

<span>The result showed that the crystal violet has lesser amount than NaOH. Thus, the limiting reactant in this chemical reaction is crystal violet and the excess reactant is NaOH.</span>

3 0

2 years ago

The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl),

choli [55]

Answer:

The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl), and calcium chloride (CaCl2) (not in that order). Which boiling point corresponds to calcium chloride?

A. 101.53° C

B. 100.00° C

C. 101.02° C

D. 100.51° C

Which of the following solutions will have the lowest freezing point?

A. 1.0 mol/kg sucrose (C12H22O11)

B. 1.0 mol/kg lithium chloride (LiCl)

C. 1.0 mol/kg sodium phosphide (Na3P)

D. 1.0 mol/kg magnesium fluoride (MgF2)

Which of the following compounds will be most effective in melting the ice on the roads when the air temperature is below zero?

A. sodium iodide (NaI)

B. magnesium sulfate (MgSO4)

C. potassium bromide (KBr)

D. All will be equally effective.

Four different solutions have the following vapor pressures at 100°C. Which solution will have the greatest boiling point?

A. 98.7 kPa

B. 96.3 kPa

C. 101.3 kPa

D. 100.2 kPa

Four different solutions have the following boiling points. Which boiling point corresponds to a solution with the lowest freezing point?

A. 101.2°C

B. 105.9°C

C. 102.7°C

D. 108.1°C

Explanation:

The following are the answers to the different questions:

The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl), and calcium chloride (CaCl2) (not in that order). Which boiling point corresponds to calcium chloride?

A. 101.53° C

Which of the following solutions will have the lowest freezing point?

D. 1.0 mol/kg magnesium fluoride (MgF2)

Which of the following compounds will be most effective in melting the ice on the roads when the air temperature is below zero?

A. sodium iodide (NaI)

Four different solutions have the following vapor pressures at 100°C. Which solution will have the greatest boiling point?

B. 96.3 kPa

Four different solutions have the following boiling points. Which boiling point corresponds to a solution with the lowest freezing point?

D. 108.1°C

5 0

2 years ago

A chemist reacted 12.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equation to d

Nataly_w [17]

Answer: 91.73g of NaCl

Explanation:

First, we solve for the number of moles of F2 using the ideal gas equation

V = 12L

P = 1.5 atm

T = 280K

R = 0.082atm.L/mol/K

n =?

PV = nRT

n = PV /RT

n = (1.5x12)/(0.082x280)

n = 0.784mol

Next, we convert this mole ( i.e 0.784mol) of F2 to mass

MM of F2 = 19x2 = 38g/mol

Mass conc of F2 = n x MM

= 0.784 x 38 = 29.792g

Equation for the reaction is given below

F2 + 2NaCl —> 2NaF + Cl2

Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol

Mass conc. of NaCl from the equation = 2 x 58.5 = 117g

Next, we find the mass of NaCl that reacted with 29.792g of F2.

From the equation,

38g of F2 redacted with 117g of NaCl.

Therefore, 29.792g of F2 will react with Xg of NaCl i.e

Xg of NaCl = (29.792 x 117)/38

= 91.73g

Therefore, 91.73g of NaCl reacted with f2

3 0

2 years ago

Starting with 1.5052g of BaCl2•2H2O and excessH2SO4, how many grams of BaSO4 can be formed?

muminat

1.5052g BaCl2.2H2O => 1.5052g / 274.25 g/mol = 0.0054884 mol
=> 0.0054884 mol Ba
<span>This means that at most 0.0054884 mol BaSO4 can form since Ba is the limiting reagent. </span>
<span>0.0054884 mol BaSO4 => 0.0054884 mol * 233.39 g/mol = 1.2809 g BaSO4</span>

6 0

2 years ago

Read 2 more answers

A 0.941 gram piece of magnesium metal is heated and reacts with oxygen. the resulting oxide weighed 1.560 grams. determine the p

Whitepunk [10]

Given:

magnesium = 0.941 gram piece

Magnesium oxide= 1.560 grams

Formula:

(magnesium / magnesium oxide) x 100 = % Mg

100% - % Mg = percent composition of each element

Solution:

(0.941g Mg) / (1.560g MgO)

= 0.603

= 60.3% Mg

100% - 60.3%

= 39.7%

39.7% is the percentage composition of each element.

5 0

2 years ago