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Ierofanga [76]
2 years ago
13

A graduated cylinder is filled with 34.4 mL of water. A toy Car that weighs 43.672-g is Gently placed in the cylinder and the wa

ter level rises to 39.0 mL. What Is the density of the toy car
Chemistry
1 answer:
Thepotemich [5.8K]2 years ago
8 0

Answer:9.49g/mL

Explanation:

Mass of toy = 43.672g

Volume of water = 34.4mL

Volume of toy + volume of water = 39mL

Volume of toy = 39 — 34.4 = 4.6mL

Density = Mass /volume

Density = 43.672/4.6 = 9.49g/mL

You might be interested in
6K + B2O3 → 3K2O + 2B
atroni [7]

Answer:

104.84 moles

Explanation:

Given data:

Moles of Boron produced = ?

Mass of B₂O₃ = 3650 g

Solution:

Chemical equation:

6K + B₂O₃    →    3K₂O + 2B

Number of moles of B₂O₃:

Number of moles = mass/ molar mass

Number of moles = 3650 g/ 69.63 g/mol

Number of moles = 52.42 mol

Now we will compare the moles of  B₂O₃ with B from balance chemical equation:

                 B₂O₃          :          B

                    1              :          2

                52.42         :        2×52.42 = 104.84

Thus from 3650 g of  B₂O₃  104.84 moles of boron will produced.

6 0
2 years ago
A 0.2-mm-thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contain
krek1111 [17]

Answer:

- 0.0249% Sb/cm

-1.2465 * 10^9 \frac{atoms}{cm^3.cm}

Explanation:

Given that:

One surface contains 1 Sb atom per  10⁸  Si atoms and the other surface contains 500 Sb atoms per  10⁸ Si atoms.

The concentration gradient in atomic percent (%) Sb  per cm can be calculated as follows:

The difference in concentration = \delta_c

The distance \delta_x = 0.2-mm = 0.02 cm

Now, the concentration of silicon at one surface containing  1 Sb atom per 10⁸ silicon atoms and at the outer surface that has 500 Sb atom per   10⁸ silicon atoms can be calculated as follows:

\frac{\delta_c}{\delta_c} = \frac{(1/10^8 -500/10^8)}{0.02cm} *100%

= - 0.0249% Sb/cm

b) The concentration (c_1) of Sb in atom/cm³ for the surface of 1 Sb atoms can be calculated by using the formula:

c_1 = \frac{(8 si atoms/unit cells)(1/10^3)}{(lattice parameter)^3/unit cell}

Lattice parameter = 5.4307 Å;  To cm ; we have

= 5.4307A^0* \frac{10^{-8}cm}{ A^0}

c_1 = \frac{(8 si atoms/unit cells)(1/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

= 0.00499*10^{17}atoms/cm^3

The concentration (c_2) of Sb in atom/cm³ for the surface of 500 Sb can be calculated as follows:

c_1 = \frac{(8 si atoms/unit cells)(500/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

   =  \frac{4*10^{-3}}{1.601*10^{-22}}

   = 2.4938*10^{17}atoms/cm^3

Finally, to calculate the concentration gradient

(\frac{\delta _c}{\delta_ x}) = \frac{c_1-c_2}{\delta_x}

(\frac{\delta _c}{\delta_ x}) = \frac{0.00499*10^{17}-2.493*10^{17}}{0.02}

= -1.2465 * 10^9 \frac{atoms}{cm^3.cm}

8 0
2 years ago
The percent composition by mass of a compound is 76.0% c, 12.8% h, and 11.2% o. the molar mass of this compound is 284.5 g/mol.
Leokris [45]
You have a few steps to solve this one. First, we'll find the molar mass by percentage of each element in the molecule. Then, we'll divide each of those relative masses by the atomic mass of each element. The number of times the mass divides into the relative mass is the number of atoms of that element in the molecule:

C: 284.5 x .76 = 216.22
H: 284.5 x .128= 36.416
O: 284.5 x .112 = 31.864.

Now we divide out each element's atomic mass (from the periodic table). it's okay if they're approximated from the decimal answer.
C: 216.22 ÷ 12.011 ≈ 18
H: 36.416 ÷ 1.008 ≈36
O: 31.864 ÷ 15.999 ≈ 2

Therefore, the molecular formula is C18H36O2. 

The empirical formula would be found by dividing out all factors of those subscript numbers. In our case, all of them can be divided by 2. The empirical formula would be C9H18O




7 0
2 years ago
Describe two shortcomings of the pennium model for isotopes
umka21 [38]

let me know when u find out plz because i would like to know as well its one of my chemistry qustions in an assiment. :)

3 0
2 years ago
The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ΔH, ΔS, a
Nostrana [21]

The question is incomplete, the complete question is;

The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ∆H, ∆S, and ∆G for the boiling process at this temperature.

A. ∆H > 0, ∆S > 0, ∆G < 0

B. ∆H > 0, ∆S > 0, ∆G > 0

C. ∆H > 0, ∆S < 0, ∆G < 0

D. ∆H < 0, ∆S > 0, ∆G > 0

E. ∆H < 0, ∆S < 0, ∆G > 0

Answer:

A. ∆H > 0, ∆S > 0, ∆G < 0

Explanation:

During boiling, a liquid is converted to vapour. This is a phase change for which heat is absorbed because energy must be taken in to break the intermolecular bonds in the liquid before it can be converted to a gas. Hence ∆H>0

Secondly, a phase change from liquid to gas leads to an increase in entropy hence ∆S>0.

Thirdly, the process is spontaneous. For every spontaneous process ∆G<0

3 0
2 years ago
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