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3 years ago

What is the mass in grams of 1.24 mol of water

1 answer:

4 0

Mass of H2O=2.01568g+15.9994g=18.01508gmol H2O.
Step 2: Multiply 1.24 mol H2O with the molecular mass. 1.24 mol H2O x 18.01508gmol H2O = 22.3 g H2O.

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Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction

inessss [21]

Answer:

Explanation:

2 NH4I + Pb(NO3)2 = PbI2 + 2 NH4NO3

2 mol 1 mol

x ml 751 ml

0.550 M 0.380 M

0.380 mol/1000 ml x 751 ml = 0.285 mol Pb(NO3)2

0.285 mol Pb(NO3)2 x 2 mol NH4I/1 mol Pb(NO3)2=0.571 mol NH4I needed

0.571 mol NH4I/0.550 mol/1000 ml=1038 ml

6 0

3 years ago

De acuerdo con la Ley de la Conservación de la Materia postulada por Lavoisier, ¿cuál ecuación química está balanceada? A C2H5OH

olya-2409 [2.1K]

Answer:

C C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O

Explanation:

La ley de conservación de la materia implica que en una reacción química los átomos en los reactivos son iguales a la cantidad de átomos en los productos:

A C2H5OH(l) + O2(g) → CO2 + H2O

En los reactivos hay dos átomos de carbono pero en los productos solo 1. De esta manera, no cumple la ley de conservación.

B C2H5OH(l) + O2(g) → 2CO2 + H2O

En los reactivos hay 6 átomos de hidrógeno pero en los productos solo 2. No cumple la ley de conservación.

C C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O

En los reactivos y productos hay: 2 átomos de carbono, 6 átomos de hidrógeno y 7 átomos de oxígeno. Cumple la ley de conservación.

D C2H5OH(l) + O2(g) → 2CO2 + 3H2O

En los reactivos hay 3 átomos de oxígeno y en los productos 7. No cumple la ley de conservación.

E 2C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O

En los reactivos hay 4 átomos de carbono y en los productos solo dos. No cumple la ley de conservación.

6 0

3 years ago

The equilibrium constant for the reaction 2x(g)+y(g)=2z(g) is 2.25 . what would be the concentration of y at equilibrium with 2

Troyanec [42]

$[\text{Y}] \approx0.337\;\text{mol}\cdot\text{dm}^{-3}$ at equilibrium.

<h3>Explanation</h3>

Concentration for each of the species:

$[\text{X}] = \dfrac{n}{V} = 2\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Y}] = \dfrac{n}{V} = 0\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Z}] = \dfrac{n}{V} = 3\; \text{mol}\cdot \text{dm}^{-3}$ .

There was no Y to start with; its concentration could only have increased. Let the change in $[\text{Y}]$ be $+x \; \text{mol}\cdot \text{dm}^{-3}$ .

Make a $\textbf{RICE}$ table.

Two moles of X will be produced and two moles of Z consumed for every one mole of Y produced. As a result, the <em>change</em> in $[\text{X}]$ will be $+2\;x \; \text{mol}\cdot \text{dm}^{-3}$ and the <em>change</em> in $[\text{Z}]$ will be $-2\;x \; \text{mol}\cdot \text{dm}^{-3}$ .

$\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& & &&&\end{array}$ .

Add the value in the C row to the I row:

What's the equation of $K_c$ for this reaction? Raise the concentration of each species to its coefficient. Products go to the numerator and reactants are on the denominator.

$K_c = \dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]}$ .

$K_c = 2.25$ . As a result,

$\dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]} = \dfrac{(3-2x)^{2}}{(2+2x)^{2} \cdot x} = K_c = 2.25$ .

$(3-2\;x)^{2}= 2.25 \cdot(2+2\;x)^{2} \cdot x\\4\;x^{2} - 12 \;x + 9 = 2.25 \;(4\;x^{3} + 8 \;x^{2} + 4 \;x)\\4\;x^{2} - 12\;x + 9 = 9 \;x^{3} + 18\;x^{2} + 9\;x\\9\;x^{3} + 14\;x^{2} + 21\;x - 9 = 0$ .

The degree of this polynomial is three. Plot the equation $y = 9\;x^{3} + 14\;x^{2} + 21\;x - 9$ on a graph and look for any zeros. There's only one zero at $x \approx 0.337$ . All three concentrations end up greater than zero.

Hence the equilibrium concentration of Y: $0.337\;\text{mol}\cdot\text{dm}^{-3}$ .

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3 years ago

How are fireworks made ?

Nadusha1986 [10]

Aerial fireworks are usually manufactured as a shell that is made up of four parts. The container consists of pasted paper. The fuse allows the shell to reach the desired altitude before exploding. A bursting charge made of black powder (like a firecracker) is at the center of the shell.

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3 years ago

Please answer this question

xz_007 [3.2K]

Answer:

ddpgzgPtflzLfzfzlfM,ogzlgotzgppgzfPlff9_<}<®}®

mark me as a brainlist

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3 years ago