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3 years ago

Chemical equation for table salt and water

Chemistry

2 answers:

bezimeni [28]3 years ago

7 0

Answer:

Chemical equation for nacl dissolving in water

NaCl(s) H2O⇌ Na+(aq)+Cl−(aq) .

Explanation:

Olenka [21]3 years ago

4 0

Answer:

sodium chloride—is NaCl.

Explanation:

Table salt is an ionic compound, which breaks into its component ions or dissociates in water. These ions are Na+ and Cl-. The sodium and chlorine atoms are present in equal amounts (1:1 ratio), arranged to form a cubic crystal lattice.

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Is MgCO3 organic or inorganic

schepotkina [342]

Its inorganic as MgCO3 is contains no carbon more hydrogen which is a crutial component of all organic compounds

4 0

3 years ago

Is this a redox reaction? give evidence (many points plz answer) 2Mg + CO2 → 2MgO + C

Aleonysh [2.5K]

Answer:

$\boxed{yeah}$

Explanation:

$CO_2 \: is \: reduced \: to→C \: by \: 2M_g \: (the \: reducing \: agent): while \\ 2M_g \: is \: oxydized \: to→2M_gO \: by \: CO_2 \: (the \: oxydizing \: agent)$

6 0

3 years ago

When 231. mg of a certain molecular compound X are dissolved in 65. g of benzene (CH), the freezing point of the solution is mea

Elan Coil [88]

Answer:

Molar mass X = 18.2 g/mol

Explanation:

Step 1: Data given

Mass of compound X = 231 mg = 0.231 grams

Mass of benzene = 65.0 grams

The freezing point of the solution is measured to be 4.5 °C.

Freezing point of pure benzene = 5.5 °C

The freezing point constant for benzene is 5.12 °C/m

Step 2: Calculate molality

ΔT = i*Kf*m

⇒ΔT = the freezing point depression = 5.5 °C - 4.5 °C = 1.0 °C

⇒i = the Van't hoff factor = 1

⇒Kf = the freezing point depression constant for benzene = 5.12 °C/m

⇒m = the molality = moles X / mass benzene

m = 1.0 / 5.12 °C/m

m = 0.1953 molal

Step 3: Calculate moles X

Moles X = molality * mass benzene

Moles X = 0.1953 molal * 0.065 kg

Moles X = 0.0127 moles

Step 4: Calculate molar mass X

Molar mass X = mass / moles

Molar mass X = 0.231 grams / 0.0127 moles

Molar mass X = 18.2 g/mol

5 0

3 years ago

A gas that was cooled to 200 Kelvin has a volume of 65.8 L. If its initial volume was 132.4 L, what was its initial temperature?

Mandarinka [93]

Answer:

Initial temperature, T1 = 99.4 Kelvin

Explanation:

Given the following data;

Initial volume, V1 = 65.8 Litres
Final temperature, T2 = 200 Kelvin
Final volume, V2 = 132.4 Litres

To find the initial temperature (T1), we would use Charles' law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles' law is given by the formula;

$\frac {V}{T} = K$