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-BARSIC- [3]
3 years ago
14

What is the situation when I might travel high velocity but low acceleration

Chemistry
1 answer:
Anna007 [38]3 years ago
6 0
Cruising at 35,000 feet in an airliner, straight toward the east, 
at 500 miles per hour.  

Since your speed is not changing and you're moving in a straight line,
your acceleration is zero.
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Classify the following as either solutions or colloids. If a colloid, name the type of colloid and identify both the dispersed a
aev [14]

Answer:

a. glucose in water( solution)

b. smoke in air (colloids)

c. carbon dioxide in air (solution)

d. milk( colloids)

Explanation:

A solution is said to be formed when a solute dissolves in a solvent to form a homogeneous mixture. The solute particles are less than 10^-9m in size. Familiar solutions are those where the solute are dissolved in a liquid solvent. When the liquid water, the solution is known as an aqueous solution. A typical example is (glucose in water). In some other cases, the apparent solution of a solute in a solvent is accompanied by a chemical reaction and this is often known as a chemical reaction. A typical example is (carbon dioxide in air).

Colloids are also known as false solutions. Here, the individual solute particles are larger than the particles of the true solution, but not large enough to be seen by the naked eye. When a light beam is placed beside a beaker containing a colloid, the light rays of the beam can be clearly seen. This shows that it exhibits the Tyndall effect while a solution dosent exhibit such.

In a colloid, the liquid solvent is more appropriately know as the DISPERSION medium while the solid solute particles constitute the DISPERSED substance. This can either be solid, liquid or gas.

For example:

--> smoke in air : Dispersion medium is gas while the dispersed substance is solid.

--> milk: Dispersion medium is liquid while the dispersed substance is liquid.

7 0
3 years ago
PLS HELPP DUE TODAY NEED DONE
Zarrin [17]

Answer:

Explanation:

Each coil increases it by a multiple of 100.

=> 50 | 3 | <u><em>15,000</em></u>

=> 100 | 3 | <u><em>30,000</em></u>

=> 150 | 3 | <u><em>45,000</em></u>

3 0
2 years ago
Read 2 more answers
The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C
Kipish [7]

Answer: 72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Putting in the values we get:

\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

According to stoichiometry:

1 mole of C_3H_8 requires 5 moles of oxygen

0.45 moles of C_3H_8 require= \frac{5}{1}\times 0.45=2.25 moles of oxygen

Mass of O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g

72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

7 0
3 years ago
A mineral consisted of 29.4% calcium, 23.5% sulfur and 47.1% oxygen. What is the empirical formula?
inessss [21]
Step  one  calculate  the  moles  of  each  element
that  is  moles= %  composition/molar  mass
molar  mass  of    Ca  =  40g/mol,     S=  32  g/mol ,   O=  16 g/mol

moles  of      Ca = 29.4 /40g/mol=0.735 moles,      S= 23.5/32  =0.734 moles,  O= 47.1/16= 2.94   moles

calculate  the mole  ratio  by  dividing   each  mole with  smallest  mole   that   is  0.734
Ca=  0.735/0.734= 1,      S=  0.734/0.734 =1,   O  = 2.94/  0.734= 4
therefore  the  emipical  formula  =  CaSO4
5 0
3 years ago
Calculate number of moles in 86.4 g of Ni
Ira Lisetskai [31]
Number of moles = mass of Ni /molecular mass of Ni

mass of nickel = 86.4 g
molecular mass of nickel = 58.69
number of moles of Ni in 86.4 g
=86.4/58.69
=1.472 mol  
(rounded to four significant figures instead of three because the first digit of the answer starts with a 1).
4 0
3 years ago
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