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avanturin [10]
2 years ago
12

You are given two unknown solutions, A and B. One of the solutions contains water and table salt, and the other contains water a

nd table sugar. Without any testing, using the data table and your knowledge of chemical bonds, you are able to determine which solution contains sugar and which contains salt. Which solution is which? A) A contains sugar and B contains salt, because salt is ionic and a good conductor of electricity. B) A contains salt and B contains sugar, because sugar is ionic and is a good conductor of electricity. C) A contains salt and B contains sugar, because salt is covalent and is poor conductor of electricity. D) A contains sugar and B contains salt, because sugar is covalent and is a good conductor of electricity.
Chemistry
2 answers:
s2008m [1.1K]2 years ago
7 0
It would be <span>A: A contains sugar and B contains salt, because salt is ionic and a good conductor of electricity.</span>
Lelechka [254]2 years ago
5 0

The answer is: A) A contains sugar and B contains salt, because salt is ionic and a good conductor of electricity.

Ionic compounds conduct electricity, because it dissociate in ions and ions conduct electicity.

Ionic substances dissociate in water on cations and anions.

For example, dissociation of sodium chloride in water: NaCl(aq) → Na⁺(aq) + Cl⁻(aq).

Ionic compounds (most of the salts, strong acids and bases) dissolve in water because an ion-dipole interactions.

An ion-dipole is electrostatic interaction between a charged ion (cations and anions) and a molecule that has a dipole (in this example water).  

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The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of C3H8 is 44.1 g/mol. What mass of O2, in grams, is required t
Ira Lisetskai [31]
The reaction formula of this is C3H8 + 5O2 --> 3CO2 + 4H2O. The ratio of mole number of C3H8 and O2 is 1:5. 0.025g equals to 0.025/44.1=0.00057 mole. So the mass of O2 is 0.00057*5*32=0.0912 g.
6 0
2 years ago
How many d electrons (n of dn) are in the central metal ion in the species below? (a) [ru(nh3)5cl]so4 6 d electrons (b) na2[os(c
Georgia [21]
For a) [Ru(NH₃)₅Cl]SO₄
Ru configuration = d⁶s²
In this complex Ru oxidation number is +3
Ru³⁺ configuration = d⁵
number of d^{n} electrons = 5

For b) Na₂[Os(CN)₆]
Os configuration = d⁶s²
In this complex Os oxidation number is +4
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number of d^{n} electrons = 4
4 0
2 years ago
Read 2 more answers
Ammonia (nh3(g), hf = –46.19 kj/mol) reacts with hydrogen chloride (hcl(g), hf = –92.30 kj/mol) to form ammonium chloride (nh4cl
steposvetlana [31]

Answer: The \Delta H_{rxn} for the given chemical reaction is -175.51 kJ/mol

Explanation: Enthalpy change of the reaction is defined as the amount of heat released or absorbed in a given chemical reaction.

Mathematically,

\Delta H_{rxn}=\Delta H_f_{(products)}-\Delta H_f_{(reactants)}

We are given a chemical reaction. The reaction follows:

NH_3(g)+HCl(g)\rightarrow NH_4Cl(s)

H_f_{(NH_3)}=-46.19kJ/mol

H_f_{(HCl)}=-92.30kJ/mol

H_f_{(NH_4Cl)}=-314.4kJ/mol

Enthalpy change for the reaction of he given chemical reaction is given by:

\Delta H_{rxn}=H_f_{(NH_4Cl)}-(H_f_{(NH_3)}+H_f_{(HCl)})

Putting the values in above equation, we get

\Delta H_{rxn}=-314.4-(-92.30-46.19)kJ/mol

\Delta H_{rxn}=-175.51kJ/mol

8 0
2 years ago
Read 2 more answers
A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water
lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

Q_{water} -Q_{sample} = 0

Q_{water} = Q_{sample}

Where:

Q_{water} - Heat received by water, measured in joules.

Q_{sample} - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)

Where:

m_{s}, m_{w} - Mass of water and the sample of aluminium, measured in kilograms.

c_{p,s}, c_{p,w} - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

T_{s}, T_{w} - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

T - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}

T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}

Given that m_{s} = 0.380\,kg, m_{w} = 2.40\,kg, c_{p,s} = 910\,\frac{J}{kg\cdot K}, c_{p,w} = 4186\,\frac{J}{kg\cdot K}, T_{s} = 378\,K and T_{w} = 273\,K, the final temperature of the system is:

T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}

T = 276.494\,K

The equilibrium temperature of the system is 276.494 Kelvin.

7 0
2 years ago
Red blood cells are placed in a solution and neither hemolysis nor crenation occurs. therefore the solution is
Len [333]

The answer is isotonic solution. These are solutions where the solute concentration in the solution and inside the cells are levelled and consequently water flows consistently. When red blood cells are positioned in an isotonic solution the cells would always stay the same.

3 0
2 years ago
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