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den301095 [7]
3 years ago
10

How do gamma rays differ from a microwave

Chemistry
1 answer:
Alex_Xolod [135]3 years ago
5 0
The difference is the photon energy which the waves use Gamma rays have the highest photon energy and microwaves have a lower photon energy,
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What equation in the blast furnace extraction of iron is not a redox reaction?
Misha Larkins [42]

The decomposition of limestone CaCO3 > CaO + CO2

8 0
3 years ago
What volume (in mililiters) of oxygen gas is required to react with 4.03 g of my at stp
goldenfox [79]

You must use 1880 mL of O₂ to react with 4.03 g Mg.

A_r: 24.305

         2Mg + O₂ ⟶ 2MgO

<em>Moles of Mg</em> = 4.03 g Mg × (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg

<em>Moles of O₂</em> = 0.1658 mol Mg × (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂

STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of <em>22.71 L</em>.

<em>Volume of O₂</em> = 0.082 90 mol O₂ × (22.71 L O₂/1 mol O₂) = 1.88 L = 1880  mL

4 0
3 years ago
HELP WITH A THESIS STATEMENT
lina2011 [118]

The second thesis statement is perfect. It supports the claim and presents main idea.

4 0
3 years ago
Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o
Gnom [1K]

Answer: The empirical formula for the given compound is C_5H_7N

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

C_xH_yN_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of CO_2=21.363mg=21.363\times 10^3g=21363g

Mass of H_2O=6.125g=6.125\times 10^3g=6125g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, \frac{12}{44}\times 21363=5826.27g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, \frac{2}{18}\times 6125=680.55 of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = \frac{485.52}{97.73}=4.96\approx 5

For Hydrogen  = \frac{680.55}{97.73}=6.96\approx 7

For Nitrogen = \frac{97.73}{97.73}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is C_5H_7N_1=C_5H_7N

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
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