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3 years ago

Propene. What does the name of this hydrocarbon tell you about it’s molecular structure?

Chemistry

1 answer:

givi [52]3 years ago

8 0

The answer is c.

There are 3 carbon atoms and six hydrogen atoms in a double-bond linear chain.

i.e.
H2C=CH-CH3

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Should we only write condensed formula in chemical equations (for lessons like carbon and compounds). What did your teachers say

Firlakuza [10]

Answer:

Here's what I get.

Explanation:

If your teachers don't ask for a specific type of formula, a condensed structural formula should be OK.
If they ask specifically for a structural formula or a bond-line formula, that is what you must give.

Bottom line: ask your teachers in advance what they expect.

7 0

3 years ago

When humans burn fossil fuels, most of the carbon quickly enters the_______

Len [333]

Answer:

Atmosphere.

Explanation:

Carbon moves from fossil fuels to the atmosphere when fuels are burned. When humans burn fossil fuels to power factories, power plants, cars and trucks, most of the carbon quickly enters the atmosphere as carbon dioxide gas.

8 0

3 years ago

Read 2 more answers

Cl2 and N2 react according to the following equation 3Cl2(g) + N2(g) rightarrow 2NCl3(g) If 4 L of a stoichiometric mixture of c

melamori03 [73]

Answer:

Volume of NCl3 is 3L

Explanation:

Avogadro states: All gases at the same volume under temperature and pressure constant have the same number of moles.

The chemical equation is:

3Cl2(g) + N2(g) → 2NCl3(g)

Where 3 moles of chlorine reacts with 1 mole of nitrogen to produce 2 moles of NCl3.

But using Avogadros law we can say:

3L of chlorine and 1L of nitrogen produce 2L of Nitrogen trichloride.

3L of chlorine and 1L of nitrogen: 4L (The stoichiometric mixture)

That means, volume of NCl3 produced is 3L

8 0

3 years ago

Balance the equations

slavikrds [6]

4. 4 C + S8 = 4 CS2

5. H2 + O2 = H2O2 is already balanced.

6. 4 Na + O2 = 2 Na2O

7 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:

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

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

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

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

_________________________________
_________________________________

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