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46.6 grams of mercury II sulfate (HgSO4) reacts with an excess of sodium Chloride (NaCl). How many grams of mercury II chloride

slega [8]

Answer:

$m_{HgCl_2}=42.7gHgCl_2$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$HgSO_4+2NaCl\rightarrow HgCl_2+Na_2SO_4$

In such a way, the mercury II sulfate (molar mass 296.65g/mol) is in a 1:1 molar ratio with the mercury II chloride (molar mass 271.52g/mol), for that reason the stoichiometry to find mass in grams of mercury II chloride turns out:

$m_{HgCl_2}=46.6gHgSO_4*\frac{1molHgSO_4}{296.65 gHgSO_4}*\frac{1molHgCl_2}{1molHgSO_4} *\frac{271.52gHgCl_2}{1molHgCl_2} \\\\m_{HgCl_2}=42.7gHgCl_2$

Best regards.

3 0

2 years ago

If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×

Elodia [21]

Answer:

2.30 × 10⁻⁶ M

Explanation:

Step 1: Given data

Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M

Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³

Step 2: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂

We will use the following expression.

Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²

[OH⁻] = 2.30 × 10⁻⁶ M

3 0

3 years ago

How many moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas?

tino4ka555 [31]

Taking into account the reaction stoichiometry, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CO₂ + 4 H₄ → CH₄ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CO₂: 1 mole
H₄: 4 moles
CH₄: 1 mole
H₂O: 2 moles

<h3>Moles of CH₄ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?

$moles of CH_{4} =\frac{85.1 moles of CO_{2}x4 moles of CH_{4} }{1 moles of CO_{2}}$

<u><em>moles of CH₄= 340.4 moles</em></u>

Then, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

Learn more about the reaction stoichiometry:

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6 0

1 year ago

A compound of mercury and oxygen is heated in order to decompose the compound. A 4.08 grams sample of mercury oxide upon heating

arlik [135]

Answer:

HgO (empirical formula)

Explanation:

4.08 - 3.78 = 0.3g (oxygen)

$(\frac{4.08}{201}) \: \: \: (\frac{0.3}{16} )$

0.02 : 0.02

0.02/0.02 : 0.02/0.02

1 : 1 (ratio)

HgO ( empirical formula)

2HgO ----> 2Hg + O2 ( your equation correct)

7 0

1 year ago

In a science lab, a student heats up a chemical from 10 °C to 25 °C which requires thermal energy of 30000 J. If mass of the obj

olganol [36]

Answer:

The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)

Explanation:

Step 1: Data given

Initial temperature = 10.0 °C

Final temperature = 25.0 °C

Energy required = 30000 J

Mass of the object = 40.0 grams

Step 2: Calculate the specific heat capacity of the object

Q = m* c * ΔT

⇒With Q = the heat required = 30000 J

⇒with m = the mass of the object = 40.0 grams

⇒with c = the specific heat capacity of the object = TO BE DETERMINED

⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C

30000 J = 40.0 g * c * 15.0 °C

c = 30000 J / (40.0 g * 15.0 °C)

c = 50 J/g°C

The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)

3 0

2 years ago