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garri49 [273]
4 years ago
15

An element has three naturally occurring isotopes. ten percent (.10) occurs as 55^x, fifteen percent (.15) occurs as isotope 56^

x, and seventy-five percent (.75) occurs as 57^x. calculate the weighted atomic mass of element x to the nearest tenth
Chemistry
1 answer:
Leviafan [203]4 years ago
5 0
The average atomic weight is, from the name itself, the average weight of all its naturally occurring isotopes. All you have to do is multiple the abundance of each isotope with its individual mass, then add them altogether.

Mass = (0.10*55)+(0.15*56)+(.75*57)
<em>Mass = 56.65 amu</em>
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Based on reactivity, which of the following elements can replace aluminum (Al) in a compound during a single replacement reactio
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3 years ago
How many electrons would be found in the Ion who’s symbol is I-
Vesnalui [34]

Answer:

54

Explanation:

Given symbol of the element:

                   I⁻

Number of electrons found in an ion with the symbol:

  This is a iodine ion:

         For an atom of iodine:

                   Electrons  = 53

                   Protons  = 53

                   Neutrons  = 74

An ion of iodine is one that has lost or gained electrons.

For this one, we have a negatively charged ion which implies that the number of electrons is 1 more than that of the protons.

  So, number of electrons  = 53 + 1  = 54

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6 0
3 years ago
If 335 g water at 35.5C loses 5750 J of heat, what is the final temperature of the water?
White raven [17]

Answer:

The final temperature is  39.58 degree Celsius

Explanation:

As we know

Q = m * c * change in temperature

Specific heat of water (c) = 4.2 joules per gram per Celsius degree

Substituting the given values we get  -

5750 = 335 * 4.2 * (X - 35.5)

X = 39.58 degree Celsius

3 0
3 years ago
An industrial synthesis of urea obtains 87.5 kg of urea upon reaction of 68.2 kg of ammonia with excess carbon dioxide. Determin
dolphi86 [110]

Answer:

The theoretical yield of urea = <u>120.35kg</u>

The percent yield for the reaction = <u>72.70%</u>

Explanation:

Lets calculate -

The given reaction is -

2NH_3(aq)+CO_2 →CH_4N_2O(aq)+H_2O (l)

Molar mass of urea CH_4N_2O= 60g/mole

Moles of NH_3 = \frac{62.8kg/mole}{17g/mole} (since Moles=\frac{mass  of  substance}{mass of one mole})

                     = 4011.76 moles

Moles of CO_2 = \frac{105kg}{44g/mole}

                = \frac{105000g}{44g/mole}

                = 2386.36 moles

Theoritically , moles of NH_3 required = double the moles of CO_2

    but , 4011.76 , the limiting reagent is NH_3

Theoritical moles of urea obtained = \frac{1 mole CH_4N_2O}{2mole NH_3}\times4011.76 mole NH_3

                                                      = 2005.88mole CH_4N_2O

Mass of 2005.88 mole of CH_4N_2O =2005.88 mole \times\frac{60g CH_4N_2O}{1mole CH_4N_2O}

                                                     = 120352.8g

                                                     120352.8g\times \frac{1kg}{1000g}

                                                     = 120.35kg

Therefore , theroritical yeild of urea = 120.35kg

Now , Percent yeild = \frac{87.5kg}{120.35kg}\times100

                                 72.70%

Thus , the percent yeild for the reaction is 72.70%

8 0
3 years ago
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