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Alenkasestr [34]
3 years ago
8

Is buoyancy a physical or chemical property?

Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
7 0
Buoyancy is considered a physical property. It is a type of physical property because it is related to the density and weight of the item, which are both physical.
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What is the mass (in grams) of 9.42 × 1024 molecules of methanol (CH3OH)?
Furkat [3]
(9.42x10^24)/(6.02x10^23)=15.6478 mol methanol
15.6478mol(32.04g/1 mol)=501.3568g methanol
8 0
4 years ago
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How many grams of H3PO4 are in 175mL of a 2.50M solution of H3PO4?
solmaris [256]
<span>(0.1875 moles)(98.004 g/mole) = 18.37575 g </span>
<span>In correct number of significant figures: 18.4 </span>
5 0
3 years ago
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How much energy is required to raise the temperature of 3 g of silver from 15°C to 20°C? (Refer to table of specific heat values
Amiraneli [1.4K]

Answer:

3.50

Explanation:

4 0
3 years ago
The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I
ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:  

Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\

Equation:  

3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\

Calculating the amount of MnCO_3

= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\=  13.1 \ kg

For point b:

Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\  of\  Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\

=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg

Equation:  

3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\

Calculating the amount of MnO_2:  

= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\=  6516 \ g \\\\=6.52 \ kg\\\\

7 0
3 years ago
CHEM HELP PLS 30 POINTS!!
chubhunter [2.5K]
1. 25 ml
2. 0.56 M
3. 3.0x 10^-2 M
Step by step explanation as follow
Use the formula (Molarity x volume) = ( molarity x volume)
5 0
3 years ago
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