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3 years ago

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I'm starting to get this feeling that I want a goth boy... Because they are starting to do some stuff to me and I need one... he

lp me please!!

2 answers:

dimulka [17.4K]3 years ago

7 0

Answer:

there trash

Explanation:

mina [271]3 years ago

5 0

Answer:

I have a PhD

sooooooooooo

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How to convert moles 3.612 x 10^23 atoms Fe

Finger [1]

Answer:

0.6 moles of Fe

Explanation:

Given data:

Number of atoms of Fe = 3.612×10²³

Number of moles = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

From atoms to mole:

3.612×10²³ atoms of Fe × 1 mole / 6.022 × 10²³ atoms

0.6 moles of Fe

6 0

3 years ago

Th muscular system and the skeletal system of work together so the body can______.

MA_775_DIABLO [31]

The answer is D because moving all of the body parts would get the heart racing and the blood pumping!

7 0

3 years ago

Read 2 more answers

Bromine (Br2) is produced by reacting HBr with O2, with water as a byproduct. The O2 is part of an air (21 mol % O2, 79 mol % N2

Karolina [17]

Answer:

The mole fractions:

$x_{HBr}=\frac{100mol}{318.5}=0.314$

$x_{Br_2}=\frac{78mol}{318.5}=0.245$

$x_{H_2O}=\frac{78mol}{318.5}=0.245$

$x_{O_2}=\frac{62.5mol}{318.5}=0.196$

Explanation:

The reaction described is:

$2 HBr (g) + 1/2 O_2 (g) \longrightarrow Br_2 (g) + H_2O (g)$

The limiting reactant is the HBr (oxygen is in excess).

a) The mass (in moles) balance for this sistem:

$n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *n_{HBr}*0.78$

(the 0.78 is because of the fractional conversion)

$n_{O_2}=\frac{ 0.5 mol O_2}{1 mol HBr} *n_{HBr}*1.25$

(the 1.25 is because of the oxygen excess)

$n_{H_2O}=\frac{ 1 mol H_2O}{1 mol Br_2} *n_{Br_2}$

There is only one degree of freedom in this sistem, you can either deffine the moles of HBr you have or the moles of Br2 you want to produce. The other variables are all linked by the equations above.

b) Base of calculation 100 mol of HBr:

$nn_{HBr}=100 mol HBr$

$n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *100mol HBr*0.78$

$n_{Br_2}=78 mol Br2$

$n_{O_2}=\frac{ 0.5 mol O_2}{1 mol HBr} *100 mol HBr*1.25$

$n_{O_2}=62.5 mol O_2$

$n_{H_2O}=n_{Br_2}= 78 mol$

$n_{total}=(78+78+100+62.5)mol= 318.5mol$

The mole fractions:

$x_{HBr}=\frac{100mol}{318.5}=0.314$

$x_{Br_2}=\frac{78mol}{318.5}=0.245$

$x_{H_2O}=\frac{78mol}{318.5}=0.245$

$x_{O_2}=\frac{62.5mol}{318.5}=0.196$

4 0

3 years ago

numbers represented by the following prefi xes: (a) mega-, (b) kilo-, (c) deci-, (d) centi-, (e) milli-, (f) micro-, (g) nano-,

Dahasolnce [82]

Explanation:

mega=10raised to the power 6

kilo=10 raised to the power 3

centi = 10 raised to the power negative 2

Milli = 10 raised to the power negative 3

nano = 10 raised to the power negative 9

pico= 10 raised to the power negative 12

micro = 10 raised to the power negative 6

5 0

3 years ago

The standard molar enthalpy of formation of NH3(g) is -45.9 kJ/mol. What is the enthalpy change if 9.51 g N2(g) and 1.96 g H2(g)

Vladimir [108]

Answer:

$\Delta H=-29.7kJ$

Explanation:

Hello!

In this case, since the undergoing chemical reaction is:

$N_2+3H_2\rightarrow 2NH_3$

We first need to identify the limiting reactant given the masses of nitrogen and hydrogen:

$n_{NH_3}^{by\ H_2}=1.96gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.647molNH_3\\\\ n_{NH_3}^{by\ N_2}=9.51gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.679molNH_3$

It means that only 0.647 moles of ammonia are yielded, so the resulting enthalpy change is:

$\Delta H=0.647molNH_3*\frac{-45.9kJ}{1molNH_3}\\\\ \Delta H=-29.7kJ$

Best regards!

3 0

3 years ago