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3 years ago

Balance this equation. If a coefficient of "1" is required, choose "blank" for that box.

2 answers:

8 0

C2H4 + O2 → CO2 + H2O

We want the same number of each element on each side.
Since there is only one compound on each side, we can start with carbon: When we double C02, we end up with 2 carbon atoms on each side.
When we double H20, we end up with 4 hydrogen atoms on each side.
The right side now contains 4 + 2 = 6 oxygen atoms, so we put a 3 in front of 02.
The result is:
C2H4 + 3O2 → 2CO2 + 2H2O

Andrej [43]3 years ago

4 0

When balancing the equation you need to know how many moles reacted and how many were formed.To balance this equation we require 1 moles of C2H4, 3moles of O2 to form 2 moles CO2 and 2 moles H2o
therefore the equation is represented as
C2H4+3O2 ->2CO2+2H2O

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About how many times longer than modern humans have hooved mammals lived on earth?

abruzzese [7]

Answer:

I think it should be about maybe the times of the dinosaurs

also I think 23 to 16 billion years

Explanation:

No explanation sorry if im wrong

8 0

3 years ago

An aqueous solution of 1.29 M ethanol, CH3CH2OH, has a density of 0.988 g/mL. The percent by mass of CH3CH2OH in the solution is

anzhelika [568]

Answer:

Explanation:

Hello, for this case, we assume that the volume of the solution is 1L, thus, the mass is given by using the density as follows:

$m_{solution}=1L*\frac{1000 mL}{1L}*\frac{0.988g}{1mL} =988g$

Now, the mass of the ethanol:

$m_{C_2H_5OH}=(1.29molC_2H_5OH/L*1L)*\frac{46gC_2H_5OH}{1molC_2H_5OH} \\m_{C_2H_5OH}=59.34g$

Finally, the by mass percent is:

$m/m=\frac{59.34g}{988g}*100\\$ %

%m=6%

Best regards.

4 0

2 years ago

A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 42 g of water (with an

professor190 [17]

Answer:

1.717 J/g °C ( third option)

Explanation:

A piece of the unknown metal dropped into water this means that Q of metal is equal to Q of the water. We write this equality as follows:

Step 1: writing the formulas:

Q = mc∆T

⇒ -Q(metal) = Q(water) Because :Metal dropped into water this means that Q of metal is equal to Q of the water.

We can write the formula different :

Mass of metal * (cmetal)(ΔT) = Mass of water *(cwater) (ΔT)

⇒ Here c is the specific heat and depends on material and phase

For this case :

mass of the metal = 68.6g

mass of the water = 42g

Specific heat of the metal = TO BE DETERMINED

Specific heat of the water = 4.184J/g °C

Initial temperature of the metal = 100 °C ⇒ Change of temperature: 52.1 - 100

Initial temperature of the water = 20°C ⇒ Change of temperature: 52.1 - 20

Step 2: Calculating specific heat of the metal

-(Mass of metal) * (cmetal)*(ΔT)) = (Mass of water) *(cwater)*(ΔT)

-68.6g (cmetal)(52.1 - 100) = 42g (4.184j/g °C) (52.1 - 20)

-68.6g *cmetal * (-47.9) = 42g (4.184j/g °C) *(32.1)

3285.94 * cmetal = 5640.87

cmetal = 5640.87 / 3285.94 = 1,71667 J/g °C

cMetal = 1.717 J/g °C

4 0

3 years ago

An exacuted glass vessel weighs 50 g when empty, 148g when filled with an liquid of density o.989/cc and 50.5 g whenfilled with

Zanzabum

MW of gas : 124.12 g/mol

<h3>Further explanation </h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of density

The unit of density can be expressed in g/cm³ or kg/m³

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

ρ = density

m = mass

v = volume

glass vessel wieight = 50 g

glass vessel + liquid = 148 ⇒ liquid = 148 - 50 =98 g

volume of glass vessel :

$\tt V=\dfrac{m}{\rho}=\dfrac{98}{0.989}=99.1~ml$

An ideal gas :

m = 50.5 - 50 = 0.5 g

P = 760 mmHg = 1 atm

T = 300 K

$\tt PV=\dfrac{mass(m)}{MW}.RT\\\\MW=\dfrac{m.RT}{PV}\\\\Mw=\dfrac{0.5\times 0.082\times 300}{1\times 0.0991}=124.12~g/mol$

3 0

3 years ago

Distilled water density 1.0 g/cm^3 propane density 0.494 g/cm^3 salt water density 1.025 g/cm^3 liquid gold density 17.31g/cm^3

I am Lyosha [343]

What are you asking?

6 0

3 years ago