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kkurt [141]
1 year ago
9

in the early studies of chemistry, scientists used properties and changed to help identify compounds. this is still done today.

if you were given the following observation about salt water, how would you classify it? “when electricity flows through a flask of salt water, bubbles form in the water. these bubbles, when collected, will burn.”
Chemistry
1 answer:
Kaylis [27]1 year ago
6 0
The salt water is a mixture because the salt and the water can be separated from each other. The dissolution of salt inside water is a physical change.  At the same time, the salt water is an electrolyte, because it conducts electricity. The electricity that was pass through the water led to the decomposition of the water; those bubbles that form inside the water represent oxygen gas which is a product of decomposition of water, therefore the electricity that was passed through the salt water has caused chemical change to occur inside the salt water.<span />
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During a titration the following data were collected. A 20.0 mL portion of solution of an unknown acid HX was titrated with 2.0
Dafna1 [17]

Answer:

The molarity of the acid HX is 6.0 M.

Explanation:

We determine the amount of moles of KOH used to neutralize the acid:

\frac{2.0moles_{KOH}}{1000ml} *60ml=0.12 moles KOH

Then, we calculate the amount of moles of acid:

0.12 moles KOH×\frac{1 mole HX}{1 moles KOH}=0.12 moles HX

The molarity of HX is:

\frac{0.12 moles HX}{20ml} *\frac{1000ml}{1l}=6.0 M

8 0
1 year ago
Suppose that 0.323 g of an unknown sulfate salt is dissolved in 50 mL of water. The solution is acidified with 6M HCl, heated, a
geniusboy [140]

Answer:

1) 41.16 % = 0.182 grams

2) The alkali cation is K+ , to form the salt K2SO4

Explanation:

Step 1: Data given

Mass of unknown sulfate salt = 0.323 grams

Volume of water = 50 mL

Molarity of HCl = 6M

Step 2: The balanced equation

SO4^2- + BaCl2 → BaSO4 + 2Cl-

Step 3: Calculate amount of SO4^2- in BaSO4

The precipitate will be BaSO4

The amount of SO4^2- in BaSO4 = (Molar mass of SO4^2-/Molar mass BaSO4)*100 %

The amount of SO4^2- in BaSO4 = (96.06 /233.38) * 100

= 41.16%

So in 0.443g of BaSO4 there will be 0.443 * 41.16 % = <u>0.182 grams</u>

<u />

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2. If it is assumed that the salt is an alkali sulfate determine the identity of the alkali cation.

The unknown sulphate salt has 0.182g of sulphate. This means the alkali cation has a weight of 0.323-0.182 = 0.141g grams

An alkali cation has a chargoe of +1; sulphate has a charge of -2

The formula will be X2SO4 (with X = the unknown alkali metal).

Calculate moles of sulphate

Moles sulphate = 0.182 grams (32.1 + 4*16)

Moles sulphate = 0.00189 moles

The moles of sulphate = 0.182/(32.1+16*4)

The moles of sulphate = 0.00189 moles

X2SO4 → 2X+ + SO4^2-

For 2 moles cation we have 1 mol anion

For 0.00189 moles anion, we have 2*0.00189 = 0.00378 moles cation

Calculate molar mass

Molar mass = mass / moles

Molar mass = 0.141 grams / 0.00378 grams

Molar mass = 37.3 g/mol

The closest alkali metal is potassium. (K2SO4 )

3 0
1 year ago
Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by No
liraira [26]

Answer:

  • Molality = 0.066 m
  • Molar mass = 608.36 g/mol

Explanation:

It seems the question is incomplete. However a web search us shows this data:

" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "

The <em>freezing-point depression</em> is expressed by:

  • ΔT=Kf * m

We put the data given by the problem and <u>solve for m</u>:

  • 2.63 °C = 40°C·kg/mol * m
  • m = 0.06575 m

For the calculation of the molar mass:<em> Molality</em> is defined as moles of solute per kilogram of solvent:

  • 0.06575 m = Moles reserpine / kg camphor
  • 25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor

We<u> calculate moles of reserpine:</u>

  • 0.06575 m = Moles reserpine / 0.025 kg camphor
  • Moles reserpine = 1.64x10⁻³ mol

Finally we use the mass of reserpine and the moles to calculate <u>the molar mass</u>:

  • 1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol

<em>Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.</em>

8 0
2 years ago
Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.
Lesechka [4]

<u>Answer:</u> The enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

For the given chemical reaction:

C(s)+2H_2(g)\rightarrow CH_4(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]

We are given:

\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ

Hence, the enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

3 0
2 years ago
What is the mass of a sample of NH3 containing 7.20 × 1024 molecules of NH3? 161 grams 187 grams 203 grams 214 grams
Shkiper50 [21]

Answer:

203 grams

Explanation:

<em>It is known that 1.0 mole of a compound contains Avogadro's number of molecules (6.022 x 10²³). </em>

<em><u>Using cross multiplication:</u></em>

1.0 mol contains → 6.022 x 10²³ molecules.

??? mol contains → 7.2 x 10²⁴ molecules.

∴ The no. of moles of (6.3 x 10²⁴ molecules) of NH₃ = (1.0 mol)(7.2 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 11.96 mol.

<em>∴ The no. of grams of NH₃ present = no. of moles x molar mass </em>= (11.96 mol)(17.0 g/mol) = <em>203.3 g ≅ 203.0 g.</em>

7 0
1 year ago
Read 2 more answers
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