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emmasim [6.3K]
3 years ago
11

(03.04 MC) What is the formula for sodium nitrite? O NaN O NaN O NaNO2 O NaNO3

Chemistry
2 answers:
charle [14.2K]3 years ago
7 0

Answer:

NaNO2 is the formula for sodium nitrite

Explanation:

Check the selected ions chart to find out more in depth.

Degger [83]3 years ago
3 0

Answer:

Formula for sodium nitrite is NaNO₂

Explanation:

Sodium nitrite is an inorganic compound with the chemical formula NaNO₂. It is a white to slightly yellowish crystalline powder that is very soluble in water and is hygroscopic.

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Bess [88]

Answer:

Explanation:

<u>Manganese (VII) ion (an anion) has the formula MnO₄⁻</u>. A polyatomic ion is an ion that is made up of more than one atom. For example, MnO₄⁻ and NH₄⁺. Since the ion provided in the question is an anion, the polyatomic ion that would react with it will have to be a cation (positively charged).

<u>The polyatomic cation that will react with MnO₄⁻ to form a neutral compound is NH₄⁺ (ammonium ion) to form NH₄MnO₄ (Ammonium permanganate).</u>

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2 years ago
I can be found in the 4th period on the periodic table.
Rainbow [258]

Vanadium (V)

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4 0
3 years ago
What determines the order of placement of the elements on the modern Periodic Table?
Pachacha [2.7K]
The order of placement of the elemnts on the modern Periodic Table is determinated by:
1)atomic number (Z=number of protons).
3 0
3 years ago
Read 2 more answers
The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

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

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

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3 years ago
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geniusboy [140]

Answer:

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Explanation:

4 0
2 years ago
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