What is when a organism reproduces with another organism

HOW TO DO STOICHIOMETRY

stellarik [79]

Answer:

1.Balance the equation.

2.Convert units of a given substance to moles.

3.Using the mole ratio, calculate the moles of substance yielded by the reaction.

4.Convert moles of wanted substance to desired units.

Explanation:

8 0

3 years ago

1.50μm2. Convert this to square meters.

Verizon [17]

1.50x10^6 m2 is the answer you're looking for

6 0

3 years ago

Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.

olasank [31]

Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

$d = \dfrac{m}{V} \to V = \dfrac{m}{d}$

$V = \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)$

The solvent volume will be:

1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

Then the mass of the solvent is:

m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{m_1}{M_1*m_2}$

$\omega = \dfrac{200}{132*0,821}$

$\omega = \dfrac{200}{108,372}$

$\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

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Another way to find the answer:

We have the following data:

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

Let's find the number of mols (n), let's see:

 $n = \dfrac{m_1}{M_1}$

$n = \dfrac{200}{132}$

$n \approx 1,5\:mol$

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{n}{m_2}$

$\omega = \dfrac{1,5}{0,8}$

$
\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

I hope this helps. =)

7 0

3 years ago

What minimum mass of H2SO4 would you need?

ollegr [7]

To do what...............

3 0

4 years ago

If 5 grams of hydrogen reacted with 40 games of oxygen to form water, how much of water would be formed?

GREYUIT [131]

2H₂ + O₂ = 2H₂O

n(H₂)=m(H₂)/M(H₂)
n(H₂)=5g/2.0g/mol=2.5 mol

n(O₂)=m(O₂)/M(O₂)
n(O₂)=40g/32.0g/mol=1.25 mol

H₂ : O₂ = 2 : 1

2.5 : 1.25 = 2 : 1

n(H₂O)=n(H₂)=2n(O₂)=2.5 mol

m(H₂O)=n(H₂O)M(H₂O)

m(H₂O)=2.5mol*18.0g/mol=45.0 g

3 0

3 years ago