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3 years ago

Cuántos moles hay en 17.5 gramod de ZnCl2

Chemistry

1 answer:

Aliun [14]3 years ago

4 0

Explanation:

Zn=65

Cl2= 35+35=70

65+70=135g

1 mole ZnCl2 = 135g

x mole = 17.5g

17.5g × 1 mole/ 135g= 0.129 moles en 17.5g de ZnCl2

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Please help me the picture is above.

Andreas93 [3]

Answer:

Explanation:

bc inorganic compoud refers to all compound that do not contain carbons.

4 0

2 years ago

Two hydrogen atoms collide in a head on collision and end up with zero kinetic energy. Each then emits a photon of 121.6 nm (n=2

Zinaida [17]

Explanation:

Expression for the kinetic energy is as follows.

K.E = $\frac{1}{2}mv^{2}$

Now, total kinetic energy will be as follows.

K.E = $2 \times \frac{1}{2}mv^{2}$ = $m \times v^{2}$

Since, this energy converts into electromagnetic radiation of wavelength 121.6 nm.

Relation between energy and photon is as follows.

Energy of photon = $\frac{hc}{\lambda}$

= $\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{121.6 \times 10^{-9}}$

= $1.63 \times 10^{-18} J$

$m \times v^{2} = 1.63 \times 10^{-18}$

v = $\sqrt{\frac{1.63 \times 10^{-18}}{1.67 \times 10^{-27}}$

= $3.12 \times 10^{4}$ m/s

Thus, we can conclude that atoms were moving at a speed of $3.12 \times 10^{4}$ m/s before the collision.

8 0

3 years ago

Equation is balanced already

mojhsa [17]

Answer:

1.) 13 g C₄H₁₀

2.) 41 g CO₂

Explanation:

To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.

Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)

Molar Mass (C₄H₁₀): 58.124 g/mol

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

Molar Mass (O₂): 2(15.998 g/mol)

Molar Mass (O₂): 31.996 g/mol

2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O

48 g O₂ 1 mole 2 moles C₄H₁₀ 58.124 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole

= 13 g C₄H₁₀

48 g O₂ 1 mole 8 moles CO₂ 44.007 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole

= 41 g CO₂

6 0

1 year ago

How many moles of argon are in 24.3 g of argon

maksim [4K]

Answer:

0.607mol

Explanation:

n(AR) = mass / molar máss

= 24.3 /40

=0.607

3 0

3 years ago

Read 2 more answers

Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou

Ulleksa [173]

Answer:

$4.36~g~XY$

Explanation:

In this case, we can start with the reaction:

$2X + Y_2~->~2XY$

If we check the reaction, we will have 2 X and Y atoms on both sides. So, the reaction is balanced. Now, the problem give to us two amounts of reagents. Therefore, we have to find the limiting reagent. The first step then is to find the moles of each compound using the molar mass:

$3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X$

$4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2$

Now, we can divide by the coefficient of each compound (given by the balanced reaction):

$\frac{0.04~mol~X}{1}=~0.04$

$\frac{0.0875~mol~Y_2}{2}=0.04375$

The smallest value is for "X", therefore this is our limiting reagent. Now, if we use the molar ratio between "X" and "XY" we can calculate the moles of XY, so:

$0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY$

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol $Y_2$ = 48 g $Y_2$ (therefore 1 mol Y = 24 g Y). With this in mind the molar mass of XY would be 85+24 = 109 g/mol. With this in mind:

$0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY$

I hope it helps!

6 0

3 years ago

Cuántos moles hay en 17.5 gramod de ZnCl2​

Cuántos moles hay en 17.5 gramod de ZnCl2