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2 years ago

Why is it wrong to assume that you are only "doing science" while you are doing experiments?

1 answer:

6 0

Science is way to broad of a subject to define a specific thing you are doing. It is the equivalent of someone asking what book you are reading and you replying, "a novel." Your experiment could also have life-changing effects on someone, so to underrate it as "doing science" would simply be inaccurate.

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37.5g of nitrogen reacts with 15.5 g of hydrogen. What mass of ammonia can be made? What is the limiting reactant?

Gennadij [26K]

Answer:

Limiting reactants or limiting reagents decide the amount of product formed and the amount of excess reagent used.

7 0

2 years ago

Explain atoms contain positively charge protons and negatively charge electrons. why does an ordinary atom have no charge?

tatiyna

An ordinary atom is balanced for instance, lithium has 3 protons and 3 electrons the positively charged proton cancels the negatively charged electron.

THINK math plus 3 and negative 3 = 0 neutral

3 0

2 years ago

A helium-filled balloon at 310.0 K and 1 atm, contains 0.05 g He, and has a volume of 1.21 L. It is placed in a freezer (T = 235

trapecia [35]

Answer : The value of $\Delta E$ of the gas is 2.79 Joules.

Explanation :

First we have to calculate the moles of helium.

$\text{Moles of helium}=\frac{\text{Mass of helium}}{\text{Molar mass of helium}}$

Molar mass of helium = 4 g/mole

$\text{Moles of helium}=\frac{0.05g}{4g/mole}=0.0125mole$

Now we have to calculate the heat.

Formula used :

$q=nc_p\Delta T\\\\q=nc_p(T_2-T_1)$

where,

q = heat

n = number of moles of helium gas = 0.0125 mole

$c_p$ = specific heat of helium = 20.8 J/mol.K

$T_1$ = initial temperature = 310.0 K

$T_2$ = final temperature = 235.0 K

Now put all the given values in the above formula, we get:

$q=nc_p(T_2-T_1)$

$q=(0.0125mole)\times (20.8J/mol.K)\times (235.0-310.0)K$

$q=-19.5J$

Now we have top calculate the work done.

Formula used :

$w=-p\Delta V\\\\w=-p(V_2-V_1)$

where,

w = work done

p = pressure of the gas = 1 atm

$V_1$ = initial volume = 1.21 L

$V_2$ = final volume = 0.99 L

Now put all the given values in the above formula, we get:

$w=-p(V_2-V_1)$

$w=-(1atm)\times (0.99-1.21)L$

$w=0.22L.artm=0.22\times 101.3J=22.29J$

conversion used : (1 L.atm = 101.3 J)

Now we have to calculate the value of $\Delta E$ of the gas.

$\Delta E=q+w$

$\Delta E=(-19.5J)+22.29J$

$\Delta E=2.79J$

Therefore, the value of $\Delta E$ of the gas is 2.79 Joules.

3 0

2 years ago

3Cu + 8HNO = 3Cu(NO3)2 +2NO + 4H2O

AysviL [449]

Taking into account the reaction stoichiometry, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 Cu+ 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Cu: 3 moles
HNO₃: 8 moles
Cu(NO₃)₂: 3 mole
NO: 2 moles
H₂O: 4 moles

The molar mass of the compounds is:

Cu: 63.55 g/mole
HNO₃: 63 g/mole
Cu(NO₃)₂: 187.55 g/mole
NO: 30 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Cu: 3 moles ×63.55 g/mole= 190.65 grams
HNO₃: 8 moles ×63 g/mole= 504 grams
Cu(NO₃)₂: 3 moles ×187.55 g/mole= 562.65 grams
NO: 2 moles ×30 g/mole= 60 grams
H₂O: 4 moles ×18 g/mole= 72 grams

<h3>Mass of Cu(NO₃)₂ produced</h3>

The following rule of three can be applied: if by reaction stoichiometry 504 grams of HNO₃ form 562.65 grams of Cu(NO₃)₂, 4.69 grams of HNO₃ form how much mass of Cu(NO₃)₂?

$mass of Cu(NO_{3} )_{2} =\frac{4.69 grams of HNO_{3} x562.65 grams of Cu(NO_{3} )_{2} }{504 grams of HNO_{3} }$