Answer:
THE SOUND TRAVELS FASTER IN SOLIDS BECAUSE ITS MOLECULES ARE CLOSE TO EACH OTHER WHILE LIQUIDS MOLECULES ARE NOT TIGHT AS OF SOLID AND IN GAS THE MOLECULES ARE FREE
Answer:
1 mole of C2H6.
Explanation:
The balanced equation for the reaction is given below:
2C2H6 + 7O2 —> 4CO2 + 6H2O
We can determine the number of mole of C2H6 that reacted to produce 2 moles of CO2 as follow:
From the balanced equation above,
2 moles of C2H6 reacted to produce 4 moles of CO2.
Therefore, Xmol of C2H6 will react to produce 2 moles of CO2 i.e
Xmol of CO2 = (2 x 2)/4
Xmol of CO2 = 1 mole.
Therefore, 1 mole of C2H6 is required to produce 2 moles of CO2.
Answer:
Option B. 4 moles of the gaseous product
Explanation:
Data obtained from the question include:
Initial volume (V1) = V
Initial number of mole (n1) = 2 moles
Final volume (V2) = 2V
Final number of mole (n2) =..?
Applying the Avogadro's law equation, we can obtain the number of mole of the gaseous product as follow:
V1/n1 = V2/n2
V/2 = 2V/n2
Cross multiply
V x n2 = 2 x 2V
Divide both side by V
n2 = (2 x 2V)/V
n2 = 2 x 2
n2 = 4 moles
Therefore, 4 moles of the gaseous product were produced.
Answer:
15.75g
Explanation:
Rearrange the formula: density = mass/volume, so that mass is the subject: mass = density × volume
Now substitute (plug in) the known values into the formula:
mass = 10.5 × 1.5 ⇒ mass = 15.75g
The mass of the silver necklace is 15.75g
Hope this helps!
Answer:
0.120M is the concentration of the solution
Explanation:
<em>Assuming the mass of sodium nitrate dissolved was 2.552g</em>
<em />
Molar concentration is an unit of concentration widely used in chemsitry defined as the moles of solute (In this case NaNO3) in 1L of solution.
To find this question we must find the moles of NaNO3 in 2.552g. With this mass and the volume (250mL = 0.250L) we can find molar concentration as follows:
<em>Moles NaNO3 -Molar mass: 84.99g/mol-</em>
2.552g * (1mol / 84.99g) = 0.0300 moles NaNO3
<em>Molar concentration:</em>
0.0300 moles NaNO3 / 0.250L =
<h3>0.120M is the concentration of the solution</h3>
