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Sergeu [11.5K]
3 years ago
7

What is the mass percent of nitrogen in ammonium carbonate, (NH4)2CO3?

Chemistry
1 answer:
Margaret [11]3 years ago
6 0

In order for you to calculate for the mass of ammonium carbonate, you need to know the molar mass of it and the nitrogen atoms in the compound. Ammonium carbonate has a molar mass of 96.08 grams per mole. There are two nitrogen atoms in ammonium carbonate which is equal to 28.02 grams per mole. Divide the molar mass of nitrogen to the ammonium carbonate, 28.02/96.08 x 100, we get 29.16wt% nitrogen.

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Is barium hydroxide Ba(OH)2 or is BaOH?
KiRa [710]

Answer:

What is the charge on the barium ion and what is the charge of the hydroxide ion.

Explanation:

To get the correct formula they have to add to zero over all.

6 0
3 years ago
If electromagnetic radiation has a frequency of 2.0×1015s−1, what is the radiation’s energy?(1 point)
Alika [10]
  • Frequency=v=2×10^15Hz
  • Energy=E

Using planks quantum theory

\\ \sf\longmapsto E=hv

\\ \sf\longmapsto E=6.626\times 10^{-34}Js\times 2\times 10^{15}s^{-1}

\\ \sf\longmapsto E=13.252\times 10^{-19}J

\\ \sf\longmapsto E=1.3\times 10^{-18}J

7 0
3 years ago
Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).
Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas         Mole Fraction      kH mol/(L*atm)

           N_2                         7.81\times 10^{-1}         6.70\times 10^{-4}

           O_2                         2.10\times 10^{-1}        1.30\times 10^{-3}

           Ar                          9.34\times 10^{-3}        1.40\times 10^{-3}

          CO_2                        3.33\times 10^{-4}        3.50\times 10^{-2}

          CH_4                       2.00\times 10^{-6}         1.40\times 10^{-3}

          H_2                          5.00\times 10^{-7}         7.80\times 10^{-4}

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}

where,

p_A = partial pressure of hydrogen gas = ?

p_T = total pressure = 0.380 atm

\chi_A = mole fraction of hydrogen gas = 5.00\times 10^{-7}

Putting values in above equation, we get:

p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{H_2}=K_H\times p_{H_2}

where,

K_H = Henry's constant = 7.80\times 10^{-4}mol/L.atm

p_{H_2} = partial pressure of hydrogen gas = 1.9\times 10^{-7}atm

Putting values in above equation, we get:

C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

4 0
3 years ago
What happens to the composition of an atom to cause it to become positively or negatively charged?
cupoosta [38]
<span>The addition and subtraction of negatively charged electrons can easily change an atom’s charge, because they perpetually spin in valence shells outside the nucleus. It is easier for a neighboring atom to share or steal an electron rather than a positively charged proton, which is found in the nucleus. It requires a strong energy input to split a proton free from other protons and neutrons. thus, the atoms lose or gain electrons from neighboring ones and become what is known as "ions". Hope it helped!</span>
6 0
4 years ago
The hydrogen-line emission spectrum includes a line at a wavelength of 434 nm. What is the energy of this radiation? (h= 6.626 x
Andrews [41]
Wavelength = 434nm = 434 x 10⁻⁹m
planck's constant = <span>h= 6.626 x 10 ⁻³⁴ J
E =?
by using the formula;
E = hc /</span>λ
value for c is 3 x 10⁸ m/s
E = (6.626 x 10 ⁻³⁴ J)(3 x 10⁸ m/s) / 434 x 10⁻⁹m
E = 1.9878 x 10⁻²⁵ / 434 x 10⁻⁹m
E = 4.58 x 10⁻¹⁹ joules

6 0
3 years ago
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