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shusha [124]
3 years ago
11

The reaction of nitrogen monoxide and its rate law are shown below. The concentration of NO is 0.025 M, and the concentration of

H2 is 0.075 M.
2NO(g)+2H2(g) -------> N2(g)+2H2O(g)

At a given temperature, the rate constant is 1.2 M-2 x s-1. What is the rate at this temperature for the given concentrations?

There are no multiple choice options. It's just (blank) M x s-1
Chemistry
1 answer:
DerKrebs [107]3 years ago
4 0
<span>The rate at this temperature for the given concentrations would be..
</span>0.000056 M × s-
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Answer these questions based on 234.04360 as the atomic mass of thorium-234. The masses for the subatomic particles are given. R
Scilla [17]

How many protons does Thorium have? 90

How many neutrons does Thorium-234 have? 144

Calculate the mass defect for the isotope thorium-234 1.85864 amu

4 0
3 years ago
Read 2 more answers
Molybdenum has a molar mass of 95.94g/mol. How many molecules of molybdenum are in 150.0 g of molybdenum
Kitty [74]
We are given the molar mass of Molybdenum as 95.94 g/mol. Also, the chemical symbol for Molybdenum is Mo. This question is asking for the amount of molecules of molybdenum in a 150.0 g sample. However, since molybdenum is a metal and it is in the form of solid molybdenum, Mo (s), it is not actual a molecule. A molecule has one or more atom bonded together. We will instead be finding the amount of atoms of Molybdenum present in the sample. To do this we use Avogadro's number, which is the amount of atoms/molecules of a substance in 1 mole of that substance.

150.0 g Mo/ 95.94 g/mol = 1.563 moles of Mo

1.563 moles Mo x 6.022 x 10²³ atoms/mole = 9.415 x 10²³ atoms Mo

Therefore, there are 9.415 x 10²³ atoms of Molybdenum in 150.0 g.
5 0
3 years ago
How many liters are in 2.75 ounces? Use the conversion factor: 1 liter = 33.814 ounces Rounded to the result to 3 significant fi
Serggg [28]
1L = 33.814 oz
xL = 2.75 oz

so it's a proportion

1L / 33.814 oz = xL / 2.75

solve for x

(1/33.814) * 2.75 = 0.0813272609 on your calculator, but it's not the answer.

the number in your problem, 2.75 oz, has 3 significant figures. so you can only round this number to 3 significant figures too.

your equipment isn't accurate enough to give a reading to 10 significant figures if that makes sense. you have to give the answer in terms of the term you use with the lowest significant figures.

so with 3 significant figures,
0.0813272609 rounds to
0.0813 L
4 0
3 years ago
elements on the right side of the periodic table differ from the elements on the left side in that elements on the right side
yarga [219]

Element  on the right side  of the periodic  table  differ from the elements on  the left side  in that elements on the <em>right  side  are non   metallic  and tends to be gases at room  temperature.</em>


<em>  </em><u>Explanation</u>

In the periodic table there element in  the right side , left side and those  which are in between.

  •  Example of element  in the right side is  fluorine  chlorine, neon, Argon   among others.
  • This element have higher  effective nuclear  charges   and  stabilize electrons more effectively.
  • there electrostatic  intermolecular  forces  are generally weak  therefore they  exist in  liquid or gaseous state.
7 0
3 years ago
A sample of food containing 27 g of fat, 48 g of carbohydrates and 20 g of protein is burned in a bomb calorimeter. In a perfect
rosijanka [135]

Answer:

38.3958 °C  

Explanation:

As,

1 gram of carbohydrates on burning gives 4 kilocalories  of energy

1 gram of protein on burning gives 4 kilocalories  of energy

1 gram of fat on burning gives 9 kilocalories of energy

Thus,

27 g of fat on burning gives 9*27 = 243 kilocalories of energy

20 g of protein on burning gives 4*20 = 80 kilocalories  of energy

48 gram of carbohydrates on burning gives 4*48 = 192 kilocalories  of energy

Total energy = 515 kilocalories

Using,

Q=m_{water}\times C_{water}\times (T_f-T_i)

Given: Volume of water = 23 L = 23×10⁻³ m³

Density=\frac{Mass}{Volume}  

Density of water= 1000 kg/m³

So, mass of the water:  

Mass\ of\ water=Density \times {Volume\ of\ water}  

Mass\ of\ water=1000 kg/m^3 \times {0.023\ m^3}  

Mass of water  = 23 kg

Initial temperature = 16°C  

Specific heat of water = 0.9998 kcal/kg°C  

515=23\times 0.9998\times (T_f-16)

Solving for final temperature as:

<u>Final temperature = 38.3958 °C  </u>

8 0
3 years ago
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