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Anuta_ua [19.1K]
2 years ago
11

If you burn a nut the same size as the cheeseball, would you get the same

Chemistry
1 answer:
tiny-mole [99]2 years ago
4 0

Answer:

No

Explanation:

The number of calories in food differentiate depending on WHAT you are eating not HOW MUCH of it that you are eating.

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Answer:

Basically, all phosphates except Sodium phosphates, Potassium phosphates and Ammonium phosphates are insoluble in water. That, of course, includes Magnesium phosphate.

Explanation:

Hope this helped!

5 0
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There are two isotopes of chlorine. below is a list of each and their mass and abundance. calculate the atomic mass of chlorine.
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The effectiveness of a chemical sanitizer can be limited by
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Chemical sanitizer effectiveness, decreases with increase in pH . And most soap and detergent are alkaline in nature, that limits its effectiveness. So they should be rinsed off completely for proper effectiveness of santizer. So improper rinsing limits the effectiveness of a sanitizer.

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7 0
3 years ago
Determine the formula for iron l oxide
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4 0
2 years ago
A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________
gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s

b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0
3 years ago
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