Atomic weight and atomic number of an element is 35 and 17 respectively

Chemistry

1 answer:

mart [117]3 years ago

7 0

Chlorine.

Chlorine is the 17th element and has a mass of 35.

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C3H8 + 5 O2 → 3 CO2 + 4 H2O<br> What is the number of atoms on each side of the equation?

Roman55 [17]

Answer: 1.2642*10²⁵ on both sides

Explanation:

First check how many moles are there on each side.
Since this is a balanaced equataion the number of moles on each side is the same thus the number of atoms is also same on both sides

There are 3 moles of carbon and 8 moles of hydrogen in C3H8
and 2 moles of oxygen in O2 but there 5 infront so 2*5 is 10
Number of moles on the right is 10+8+3 = 21

Now use Avogrado's constant

21 Moles* (6.02*10²³)/Mol
= 21*6.02*10²³

= 1.2642*10²⁵

6 0

2 years ago

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Which side does the light always start from?

jarptica [38.1K]

Answer:

in prism

it's from the rectangular reflecting surface

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2 years ago

The balanced chemical equation for the combustion of butane is: 2C2H2 + 5O2 CO2 + 2H2O 2CH4 + 5O2 2CO2 + 4H2O 2C4H10 + 13O2 8CO2

SCORPION-xisa [38]

Butane is C₄H₁₀.

$C_4H_{10} + O_2 \to CO_2 + H_2O \\ \\
\hbox{balance carbon and hydrogen on the right-hand side:} \\
C_4 H_{10} + O_2 \to 4 \ CO_2 + 5 \ H_2O \\ \\
\hbox{balance oxygen on the left-hand side:} \\
C_4 H_{10} + \frac{13}{2} \ O_2 \to 4 \ CO_2 + 5 \ H_2O \\ \\
\hbox{multiply by 2 to get rid of the fraction:} \\
2 \ C_4H_{10} + 13 \ O_2 \to 8 \ CO_2 + 10 \ H_2O$

The balanced equation is 2 C₄H₁₀ + 13 O₂ <span>→</span> 8 CO₂ + 10 H₂O.

3 0

3 years ago

Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.

Ad libitum [116K]

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^{3}$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 \times 10^{25} ions/m^{3}$

T = $30^{o}C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_{D}$ ) is as follows.

$\lambda_{D}$ = $\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

where, $n^{o}$ = concentration = $6.022 \times 10^{25} ions/m^{3}$

Hence, putting the given values into the above equation as follows.

$\lambda_{D}$ = $\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

= $\sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}$