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Levart [38]
3 years ago
12

Please help me !!! im stuck

Chemistry
2 answers:
frutty [35]3 years ago
8 0

The answer is 421.37 grams

My science teacher said it always the first 2 numbers so yea thats why its 37

Artemon [7]3 years ago
3 0
421.37 I got this question before
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PLZ HELP THIS IS DUE TODAY!!!!!!
Goshia [24]
It would be letter C, because the molecules of solids can have a little vibration in their places.
The first one is for gas, so it's wrong. The second one is for liquids. And finally, the third one is right.
5 0
3 years ago
What 2 chemicals react together during photosynthesis
Nitella [24]

Answer:

The photosynthesis chemical equation states that the reactants (carbon dioxide, water and sunlight), yield two products, glucose and oxygen gas. The single chemical equation represents the overall process of photosynthesis.

Explanation:

Definition: the process by which green plants and some other organisms use sunlight to synthesize foods from carbon dioxide and water. Photosynthesis in plants generally involves the green pigment chlorophyll and generates oxygen as a byproduct.

Hope this helps!

Can I get brainliest?

5 0
4 years ago
A kno3 solution containing 51 g of kno3 per 100.0 g of water is cooled from 40 ∘c to 0 ∘c. what will happen during cooling?
Gre4nikov [31]

Answer:

As, the temperature decreased from 40.0 °C to 0.0 °C an amount will be recrystallized and precipitated as solid crystals in the water (51.0 g - 14.0 g = 37.0 g) and 14.0 g will be dissolved in water.

Explanation:

  • Firstly, we must mention that:

The solubility of KNO₃ per 100.0 g of water at 40.0 °C = 63.0 g.

The solubility of KNO₃ per 100.0 g of water at 0.0 °C = 14.0 g.

  • So, at 40.0 °C, 51.0 g of KNO₃ will be completely dissolved in water.
  • <em>As, the temperature decreased from 40.0 °C to 0.0 °C an amount will be recrystallized and precipitated as solid crystals in the water (51.0 g - 14.0 g = 37.0 g) and 14.0 g will be dissolved in water.</em>
5 0
3 years ago
The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:
uysha [10]

Answer:

The wavelength of the line in the absorption line spectrum of hydrogen caused by transition is 657.1 nm.

Explanation:

Energy of the electron in a hydrogen atom in nth shell:

E=\frac{-R_y}{n^2}

R_y=2.178\times 10^{-18} J

Energy of an electron in n = 2:

E_2=\frac{-R_y}{2^2}=-\frac{R_y}{4}

Energy of an electron in n = 3:

E_3=\frac{-R_y}{3^2}=-\frac{R_y}{9}

Energy difference in between both the shells:

\Delta E=E_3-E_2 :

=-\frac{R_y}{9}-(-\frac{R_y}{4})=\frac{5R_y}{36}

\Delta E=\frac{5R_y}{36} =\frac{5\times 2.178\times 10^{-18} J}{36}

=3.025\times 10^{-19} J

\Delta E=3.025\times 10^{-19} J=\frac{hc}{\lambda}

\lambda =\frac{hc\times 36}{3.025\times 10^{-19} J}

=\frac{6.626\times 10^{-34} J s\times 3\times 10^8 m/s}{3.025\times 10^{-19} J}

\lambda =6.571\times 10^{-7} m=657.1 nm

1 m = 10^9 nm

The wavelength of the line in the absorption line spectrum of hydrogen caused by transition is 657.1 nm.

8 0
4 years ago
Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79
BARSIC [14]

Answer :  The mole fraction and partial pressure of CH_4,CO_2 and He gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of CH_4 = 1.79 mole

Moles of CO_2 = 1.20 mole

Moles of He = 3.71 mole

Now we have to calculate the mole fraction of CH_4,CO_2 and He gases.

\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267

and,

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179

and,

\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554

Thus, the mole fraction of CH_4,CO_2 and He gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of CH_4,CO_2 and He gases.

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 5.78 atm

X_i = mole fraction of gas

p_{CH_4}=X_{CH_4}\times p_T

p_{CH_4}=0.267\times 5.78atm=1.54atm

and,

p_{CO_2}=X_{CO_2}\times p_T

p_{CO_2}=0.179\times 5.78atm=1.03atm

and,

p_{He}=X_{He}\times p_T

p_{He}=0.554\times 5.78atm=3.20atm

Thus, the partial pressure of CH_4,CO_2 and He gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0
3 years ago
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