The main information we have to use here is the density of gold. From literature, the density of gold at room temperature is 19.32 g/cm³. To determine the mass, let's calculate the volume first. A wire is in the shape of a cylinder. Thus, the volume would be
V = πd²h/4
V = π(0.175 cm)²(1×10⁵ cm)/4
V = 2,405.28 cm³
Density = mass/volume
19.32 g/cm³ = Mass/2,405.28 cm³
Mass = 46,470 g gold wire
Answer:
0.5 g/mL----- float
1.0 g/mL---- float
2.0 g/mL-----sink
Explanation:
Those objects will float whose density is less or equal to water density due to lower mass while those object will sink whose density is greater than water because the mass of the object is greater than water. So the density of the object i. e. 0.5 g/mL and 1.0 g/mL will float in the liquid because its density is lower than water which is 1 g/mL while the density of an object i. e. 2.0 g/mL is denser than water so it will sink.
Answer:
There are 2 moles of oxygen molecules; there are 4 moles of oxygen atoms.
Explanation:
Elements in the same group<span> in the </span>periodic table<span> have </span>similar<span> chemical properties. This is because their atoms have the </span>same<span> number of electrons in the highest occupied energy level. </span>Group<span> 1 </span>elements<span> are reactive metals called the alkali metals.</span>Group<span> 0 </span>elements<span> are unreactive non-metals called the noble gases.</span>
The initial mass of sodium hydroxide is 3.3 g (answer C)
<u><em>calculation</em></u>
Step 1 : find the moles of iron (ii) hydroxide ( Fe(OH)₂
moles = mass÷ molar mass
from periodic table the molar mass of Fe(OH)₂ = 56 + [16 +1]2 = 90 g/mol
moles is therefore = 3.70 g÷ 90 g/mol = 0.041 moles
Step 2: use the mole ratio to calculate the moles of sodium hydroxide (NaOH)
from given equation NaOH : Fe(OH)₂ is 2 :1
therefore the moles of NaOH = 0.041 x 2 = 0.082 moles
Step 3: find mass of NaOH
mass = moles x molar mass
from the periodic table the molar mass of NaOH = 23 +16 +1 = 40 g/mol
mass = 0.082 moles x 40 g/mol = 3.3 g ( answer C)
