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MakcuM [25]
3 years ago
9

How many atoms are in the molecule beryllium carbonate

Chemistry
2 answers:
Vladimir79 [104]3 years ago
6 0

Answer:

1

Explanation:

g00gle

taurus [48]3 years ago
4 0
Answer is 1 you can double check
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PLEASE ANSWER ASAP AND SHOW WORK: A camping stove uses a 5.0 L propane tank that holds 68.0 moles of liquid C3H8. How large a co
seropon [69]

<u>Answer:</u> The volume of the container needed is 554.6 L

<u>Explanation:</u>

To calculate the volume of the gas, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 3.0 atm

V = Volume of the gas = ? L

T = Temperature of the gas = 25^oC=[25+273]K=298K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles of propane gas = 68.0 moles

Putting values in above equation, we get:

3.0atm\times ?L=68mol\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\V=\frac{68\times 0.0821\times 298}{3.0}=554.6L

Hence, the volume of the container needed is 554.6 L

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3 years ago
If my aunt means degree Celsius or degrees Fahrenheit
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Ozone is produced by incomplete burning of fuels combustion of sulfur containing fuel decaying organic matter photochemical oxid
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What does the graph show about the rate of temperature change over time?
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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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