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alisha [4.7K]
3 years ago
10

A double-replacement reaction takes place when aqueous Na,co, reacts with aqueous Sn(NO3), according to the following equation.

Na,co, + Sn(NO,)
You would expect one of the products of this reaction to be
20
Sn2(CO3)2
NaNO,
O Nas
OCNO,
Chemistry
1 answer:
Vladimir [108]3 years ago
8 0

Answer:

Sn(CO_3)_2 and NaNO_3.

Explanation:

Hello there!

In this case, according to the described chemical reaction between sodium carbonate and tin nitrate, we can write the undergoing chemical reaction:

2Na_2CO_3+Sn(NO_3)_4\rightarrow Sn(CO_3)_2+4NaNO_3

Thus, we notice that the products are strontium carbonate and sodium nitrate, whose formulas are Sn(CO_3)_2 and NaNO_3 respectively.

Best regards!

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Answer:

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Explanation:

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3 years ago
Calculate the standard enthalpy of formation of NOCl(g) at 25 ºC, knowing that the standard enthalpy of formation of NO(g) at th
stepan [7]

Answer:

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol

Explanation:

The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants

In this case, you have:  2 NOCl(g) → 2 NO(g) + Cl₂(g)

So, ΔH=2*H_{NO} +H_{Cl_{2} }-2*H_{NOCl}

Knowing:

  • ΔH= 75.5 kJ/mol
  • H_{NO}= 90.25 kJ/mol
  • H_{Cl_{2} }= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound  the chlorine Cl₂)
  • H_{NOCl}=?

Replacing:

75.5 kJ/mol=2* 90.25 kJ/mol + 0 - H_{NOCl}

Solving

-H_{NOCl}=75.5 kJ/mol - 2*90.25 kJ/mol

-H_{NOCl}=-105 kJ/mol

H_{NOCl}=105 kJ/mol

<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>

8 0
3 years ago
A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o
Andreas93 [3]
<h3>Answer:</h3>

2.809 L of H₂SO₄

<h3>Explanation:</h3>

Concept tested: Moles and Molarity

In this case we are give;

Mass of solid sodium hydroxide as 13.20 g

Molarity of H₂SO₄ as 0.235 M

We are required to determine the volume of H₂SO₄ required

<h3>First: We need to write the balanced equation for the reaction.</h3>
  • The reaction between NaOH and H₂SO₄ is a neutralization reaction.
  • The balanced equation for the reaction is;

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

<h3>Second: We calculate the umber of moles of NaOH used </h3>
  • Number of moles = Mass ÷ Molar mass
  • Molar mass of NaOH is 40.0 g/mol
  • Therefore;

Moles of NaOH = 13.20 g ÷ 40.0 g/mol

                          = 0.33 moles

<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>
  • From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
  • Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
  • Thus, Moles of H₂SO₄ = moles of NaOH × 2

                                    = 0.33 moles × 2

                                   = 0.66 moles of H₂SO₄

<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>
  • When given the molarity of an acid and the number of moles we can calculate the volume of the acid.
  • That is; Volume = Number of moles ÷ Molarity

In this case;

Volume of the acid = 0.66 moles ÷ 0.235 M

                                = 2.809 L

Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.

7 0
3 years ago
What is the mass of 2.49 cm of aluminum? The density of aluminum is 2.70 g/cm.
OverLord2011 [107]

Answer:

6.723

Explanation:

2.49*2.70

5 0
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