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jonny [76]
3 years ago
11

In a laboratory 0.500 moles of NAOH is completely dissolved in distilled water to form 400.0 mL of NAOH. The solution is then us

ed to titrate a solution of HNO3.
Which answer choice correctly completes the equation by filling in the formulas of the products in this titration reaction?

NaOH(aq) + HNO3(aq) —> —————- + —————-


1: NaNO2(aq) + H2O2(L)

2: NaNO3(aq) + H2O(L)

3: HNaNO3(aq) + OH(L)

4:H2NO3(aq) + NaO(L)
Chemistry
2 answers:
Eduardwww [97]3 years ago
5 0

Answer:

The answer to your question is number 2.  NaNO₃   +       H₂O

Explanation:

The reaction that is represented in this problem is an acid-base reaction. An acid-base reaction is a double displacement reaction.

The cation of a reactant attaches to the anion of the other reactant and viceversa.

Reactants      NaOH  and    HNO₃

Cations          Na⁺                 H⁺

Anions          OH⁻                NO₃⁻

Following the previous rule

Products       NaNO₃   +       H₂O

The products are a salt and water.

V125BC [204]3 years ago
4 0

Answer:

NaNO2(aq) + H2O2(l)

Explanation:

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7 0
3 years ago
Need help !!!!! ASAP
Ksivusya [100]
<h2>Hello!</h2>

The answer is:

We have that there were produced 0.120 moles of CO_{2}

n=0.120mol

<h2>Why?</h2>

We are asked to calculate the number of moles of the given gas, also, we  are given the volume, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:

PV=nRT

Where,

P is the pressure of the gas.

V is the volume of the gas.

n is the number of moles of the gas.

T is the absolute temperature of the gas (Kelvin).

R is the ideal gas constant (to work with pressure in mmHg), which is equal to:

R=62.363\frac{mmHg.L}{mol.K}

We must remember that the The Ideal Gas Law equation works with absolute temperatures (K), so, if we are given relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert it to Kelvin before we proceed to work with the equation.

We can convert from Celsius degrees to Kelvin using the following formula:

Temperature(K)=Temperature(C\°) + 273K

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Pressure=760mmHg\\Volume=2.965L\\Temperature=25.5C\°=25.5+273K=298.5K

Now, isolating the number of moles, and substituting the given information, we have:

PV=nRT

n=\frac{PV}{RT}

n=\frac{PV}{RT}

n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}

n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}\\\\n=\frac{2242mmHg.L}{18615.355\frac{mmHg.L}{mol.}}\\\\n=0.120mole

Hence, we have that there were produced 0.120 moles of CO_{2}

n=0.120mol

Have a nice day!

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Calculate the ratio of the mass ratio of SS to OO in SOSO to the mass ratio of SS to OO in SO2SO2. Consider a sample of SOSO in
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Answer:

The ratio of the mass ratio of S to O; in SO, to the mass ratio of S to O; in SO₂, is  2:1

Explanation:

According to the consideration, let us first find the ratio of S and O in both the compounds

For SO:

\frac{m_{S} }{m_{O} }= \frac{32}{16}\\\\   \frac{m_{S} }{m_{O} }= 2

Let us express it as

SO_{\frac{m_{S} }{m_{O} }} = 2

For SO₂,

Due to two oxygen atoms in the molecule, the mass of oxygen will be taken two times

\frac{m_{S} }{m_{O} }= \frac{32}{(2)(16)}\\\\   \frac{m_{S} }{m_{O} }= 1

Let us express it as

SO_2_{\frac{m_{S} }{m_{O} }}= 1

Now, for the ratio of both the above-calculated ratios,

\frac{SO_{\frac{m_{S} }{m_{O} }}}{SO_2_{\frac{m_{S} }{m_{O} }}}=\frac{2}{1}

The required ratio is 2:1

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