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balandron [24]
3 years ago
7

Which coefficients balance the following equation?

Chemistry
1 answer:
IRISSAK [1]3 years ago
7 0
Answer: D
2KClO3 = 2KCl + 3O2
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What is the highest occupied energy level for an atom of copper (Cu)? *
bonufazy [111]

Answer:

third energy level - 18

hope this helps

8 0
3 years ago
How many grams are in 2.3 x 10^-4 moles of Ca3(PO4)2
OLEGan [10]
Molar mass 

Ca₃(PO₄)₂ = 310 g/mol

1 mole -------------------> 310 g
2.3x10⁻⁴ mole ---------> ?

m = 2.3x10⁻⁴ * 310 / 1

m = 0.0713 g 

hope this helps!
3 0
3 years ago
What is the molar mass of wax (C25H52)?<br> (hydrogen- 1.01)<br> (carbon- 12.01)
garri49 [273]

Answer:

353 grams per mole

Explanation:

Candle wax is defined by the chemical formula C25H52 and has a molar mass of 353 grams per mole.

4 0
3 years ago
Read 2 more answers
The density of ammonia gas under certain condition is 0.625 g/L. Calculate its density in g/cm³.
oee [108]

Answer:

The density of Ammonia : 6.25.10⁻⁴ g/cm³

Further explanation  

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 density  

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

Density formula:  

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

ρ = density  

m = mass  

v = volume  

A common example is the water density of 1 gr/cm³  

Ammonia has a density of 0.625 g/L, then convert to g/cm³ :

\begin{gathered}\rm 1~L=1~dm^3=10^3~cm^3\\\\0.625~\dfrac{g}{L}\times \dfrac{1~L}{10^3~cm^3}\\\\\rho=\dfrac{0.625}{10^3}}\dfrac{g}{cm^3}=\boxed{6.25.10^{-4}\dfrac{g}{cm^3}}\end{gathered}

Learn more

4 0
2 years ago
Read 2 more answers
g Suppose you are titrating vinegar, which is an acetic acid solution of unknown concentration, with a sodium hydroxide solution
Elza [17]

Answer: The molar concentration of acetic acid in the vinegar is 0.539 M.

Explanation:

The formula used is:

M_1V_1=M_2V_2

where,

M_1 and V_1 are the concentration and volume of base.

M_2 and V_2 are the concentration and volume of an acid.

Given:

Molar concentration of NaOH = 0.1798 M

Volume of NaOH = 30.01 mL

Volume of acetic acid = 10.0 mL

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

M_1V_1=M_2V_2\\\\0.1798M\times 30.01mL=M_2\times 10.0mL\\\\M_2=0.539M

Thu, the molar concentration of acetic acid in the vinegar is 0.539 M.

4 0
3 years ago
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