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2 years ago

11

What type of product is formed when acids are added to some ionic compounds?

1 answer:

Andru [333]2 years ago

3 0

Answer:

<em>Your</em><em> </em><em>Answer</em><em> </em><em>is</em><em> </em><em>Option</em><em> </em><em>A</em><em> </em><em>that</em><em> </em><em>is</em><em> </em><em>Gas</em><em>.</em>

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An aqueous potassium iodate (KIO3) solution is made by dissolving 553 grams of KIO3 in sufficient water so that the final volume

liberstina [14]

Answer:

M KIO3 = 1.254 mol/L

Explanation:

molarity (M) [=] mol/L

∴ w KIO3 = 553 g

∴ mm KIO3 = 214.001 g/mol

∴ volumen sln = 2.10 L

⇒ mol KIO3 = (553 g)×(mol/210.001 g) = 2.633 mol

⇒ M KIO3 = (2.633 mol KIO3 / (2.10 L sln)

⇒ M KIO3 = 1.254 mol/L

3 0

3 years ago

Read 2 more answers

A compound distributes between benzene (solvent 1) and water (solvent 2) with a distribution coefficient, K = 2.7. If 1.0g of th

mote1985 [20]

Explanation:

The given data is as follows.

Solvent 1 = benzene, Solvent 2 = water

$K_{p}$ = 2.7, $V_{S_{2}}$ = 100 mL

$V_{S_{1}}$ = 10 mL, weight of compound = 1 g

Extract = 3

Therefore, calculate the fraction remaining as follows.

$f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}$

= $[1 + 2.7(\frac{100}{10})]^{-3}$

= $(28)^{-3}$

= $4.55 \times 10^{-5}$

Hence, weight of compound to be extracted = weight of compound - fraction remaining

= 1 - $4.55 \times 10^{-5}$

= 0.00001

or, = $1 \times 10^{-5}$

Thus, we can conclude that weight of compound that could be extracted is $1 \times 10^{-5}$ .

7 0

3 years ago

Read 2 more answers

A hot lump of 39.9 g of iron at an initial temperature of 78.1 °C is placed in 50.0 mL H 2 O initially at 25.0 °C and allowed to

Drupady [299]

Answer : The final temperature of the mixture is $29.6^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron = $0.499J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of iron = 39.9 g

$m_2$ = mass of water = $Density\times Volume=1g/mL\times 50.0mL=50.0g$

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of iron = $78.1^oC$

$T_2$ = initial temperature of water = $25.0^oC$

Now put all the given values in the above formula, we get

$(39.9g)\times (0.499J/g^oC)\times (T_f-78.1)^oC=-(50.0g)\times 4.18J/g^oC\times (T_f-25.0)^oC$

$T_f=29.6^oC$

Therefore, the final temperature of the mixture is $29.6^oC$

8 0

3 years ago

8. The standard enthalpy of formation of RbF(s) is –557.7kJ/mol and the standard enthalpy of formation of RbF(aq, 1 m) is –583.8

garri49 [273]

Explanation:

Given

The enthalpy of formation of RbF (s) is –557.7kJ/mol

The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol

The enthalpy of solution of RbF = Enthalpy of RbF (aq) - Enthalpy of formation of RbF (s)

= -583.8 - (-557.7) kJ/mol

= -26.1 kJ/mol

The enthalpy is negative which means that the temperature will rise when RbF is dissolved.

3 0

3 years ago

15g of FeCI3 is dissolved in 450 mL of solution. What is the concentration of [CI-]?

storchak [24]

The concentration of [CI-] : 0.617 M

<h3>Further explanation</h3>

FeCl₃ dissolved in 450 mL of solution(will dissociate )

Reaction

FeCl₃⇒Fe³⁺+3Cl⁻

mol FeCl₃(MW=162,2 g/mol)

$\tt \dfrac{15}{162.2}=0.0925$

mol Cl⁻ :

$\tt \dfrac{3}{1}\times 0.0925=0.2775$

molarity of Cl⁻ :

$\tt \dfrac{0.2775}{0.45}=0.617~M$

7 0

3 years ago