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2 years ago

Cual es la masa atomica del hidrogeno

1 answer:

4 0

El hidrógeno es el elemento químico de número atómico 1, representado por el símbolo H. Con una masa atómica de 1.00784 u es el más ligero de la tabla periódica de los elementos. Por lo general, se presenta en su forma molecular, formando el gas diatómico H₂ en condiciones normales.

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How many moles of carbon in 130g of C2H6

-BARSIC- [3]

7.22 moles of C2H6. Since there are 2 carbon atoms per C2H6, we must multiply the number of moles of C2H6 by 2 to get the number of moles of Carbon which is 14.4 or 14 if using two sig figs.

5 0

3 years ago

Read 2 more answers

II. The student then takes a 1.00 M stock

-BARSIC- [3]

Answer:

0.244 M.

Explanation:

From the question given above, the following data were obtained:

Molarity of stock solution (M₁) = 1 M

Volume of stock solution (V₁) = 0.305 L

Volume of diluted solution (V₂) = 1.25 L

Molarity of diluted solution (M₂) =?

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

1 × 0.305 = M₂ × 1.25

0.305 = M₂ × 1.25

Divide both side by 1.25

M₂ = 0.305 / 1.25

M₂ = 0.244 M

Thus, the molarity of the diluted solution is 0.244 M

3 0

2 years ago

The sodium atom contains 11 electrons, 11 protons, and 12 neutrons. what is the atomic mass of sodium?

Gnoma [55]

Atomic mass is the number of protons + number of neutrons. 11 protons + 12 neturons = 23

7 0

2 years ago

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$