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Ronch [10]
3 years ago
12

For the reaction 2AgNO3+Na2CrO4⟶Ag2CrO4+2NaNO3 how many grams of sodium chromate, Na2CrO4, are needed to react completely with 4

5.5 g of silver nitrate, AgNO3?
Chemistry
1 answer:
Cerrena [4.2K]3 years ago
3 0

Answer:

28.12 g Na2CrO4

Explanation:

To solve this question, we have to work with stoichiometry:

Then, if we have 45.5 g of Silver Nitrate:

45,5 g AgNo3 x \frac{1 mol AgNo3}{169.87 g AgNo3} = 0.2678 mol AgNo3\\

We know that for each Na2CrO4's mol, we need 2 AgNo3's moles, so, we can calculate the amount of AgNo3's moles as follows:

0.2678 mol AgNo3 * \frac{1 mol Na2CrO4}{2 Mol AgNO3} =0.134 mol Na2CrO4\\

and, now we can calculate the grams of sodium chromate as:

0.134 mol Na2CrO4 *\frac{209.9714 g Na2CrO4}{ molNa2CrO4} =28.12 g Na2CrO4

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sleet_krkn [62]

Answer:

\large \boxed{\text{0.091 atm }}

Explanation:

The balanced equation is

I₂(g) + Br₂(g) ⇌ 2IBr(g)

Data:

   Kc = 8.50 × 10⁻³

n(IBr) = 0.0600 mol

     V = 1.0 L

1. Calculate [IBr]

\text{[IBr]} = \dfrac{\text{0.0600 mol}}{\text{1.0 L}} = \text{0.0600 mol/L}

2. Set up an ICE table.

\begin{array}{ccccccc}\rm \text{I}_{2}& + & \text{Br}_{2} & \, \rightleftharpoons \, & \text{2IBr} &  &  \\0 & & 0 & &0.0600 & & \\+x &  & +x &   &- 2x & & \\x &   & x &   & 0.0600 - 2x & & \\\end{array}

3. Calculate [I₂]

\begin{array}{rcl}K_{\text{c}}&=&\dfrac{\text{[IBr]}^{2}} {\text{[I$_{2}$][Br]$_{2}$}}\\\\8.50 \times 10^{-2}&=&{\dfrac{(0.0600 - 2x)^{2}}{x^{2}}}& &\\\\0.2915x & = &{\dfrac{0.0600 - 2x}{x}}& &\\\\0.2915x & = &0.0600 - 2x\\\\2.2915x & = & 0.0600\\x & = & \textbf{0.026 18 mol/L}\\\end{array}\\

4. Convert the temperature to kelvins

T = (150 + 273.15) K = 423.15 K

5. Calculate p(I₂)

\begin{array}{rcl}\\pV & = & nRT\\p & = & cRT\\p & = & \text{0.026 18 mol} \cdot \text{L}^{-1}\times \text{0.082 06 L} \cdot \text{atm} \cdot \text{K}^{-1} \text{mol}^{-1} \times \text{423.15 K}\\& = & \textbf{0.91 atm}\\\end{array}\\\text{The partial pressure of iodine is $\large \boxed{\textbf{0.91 atm}}$}

6 0
3 years ago
What is pressure in relation to a gas?
labwork [276]

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Pressure is defined as force per unit area. Gas molecules are constantly colliding against the walls of the container. The pressure of the gas is the force the gas is exerting on its container.

Since temperature is defined as the average kinetic energy of the molecules of a gas then the higher the temperature, the faster the particles move.

The volume of a container refers the size if the container.

The pressure of a gas is inversely proportional to its volume according to Boyle's law. Thus implies that if the pressure of the gas goes up, the volume has to go down.

For a compound to be called an acid, it must contain H+ and H3O+ when dissolved in water.

For a compound to be called a base, the compound must dissolve in water to yield hydroxide ions.

Learn more: brainly.com/question/11543614

7 0
2 years ago
How many grams of magnesium chloride can be produced from 2.30 moles of chlorine gas reacting w excess magnesium Mg(s)+Cl2(g)-&g
alexdok [17]

Answer:

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Explanation:

6 0
3 years ago
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AlladinOne [14]

Answer:

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Explanation:

Density is a scalar quantity that is denoted by the symbol ρ (rho). It is defined as the ratio of the mass (m) of the given sample and the total volume (V) of the sample.

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According to the ideal gas law for ideal gas:

PV = nRT                                       ......equation (2)

Here, V is the volume of gas, P is the pressure of gas, T is the absolute temperature, R is Gas constant and n is the number of moles of gas

As we know,

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where m is the given mass of gas and M is the molar mass of the gas

So equation (2) can be written as:

PV = \frac{m}{M}RT

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\frac{PM}{RT}= \frac{m}{V} = \rho  

⇒ Density of an ideal gas: \rho = \frac{PM}{RT}  

⇒ <em>Density of an ideal gas: ρ ∝ molar mass of gas: M</em>

<u>Therefore, the density of the ideal gas is directly proportional to its molar mass. </u>

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4 years ago
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My science text book said that it was either diamond or gold.  Gold may not be right, but I am pretty sure diamond is.

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3 years ago
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