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OLEGan [10]
3 years ago
13

How do you know if flammability is a chemical property

Chemistry
2 answers:
Tpy6a [65]3 years ago
7 0

Answer:

When a fuel burns, it combines with oxygen int the air and changes into the substances water and carbon dioxide.

Explanation:

snow_lady [41]3 years ago
5 0
When a fuel burns, it combines with oxygen int the air and changes into the substances water and carbon dioxide. That's how Flammability is a chemical property.
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Which of these elements has the highest first ionization energy?
julia-pushkina [17]

Answer:

Cesium (Cs)

Explanation:

This eliment has an ionization energy of 3,8939

6 0
3 years ago
At 35°C and 2 atm of pressure, the volume of gas is 200 ml. If the temperature changes to
lys-0071 [83]

Answer:

V₂ = 106.5 mL

Explanation:

Given data:

Initial volume =200 mL

Initial pressure = 2 atm

Initial temperature = 35 °C (35 +273 = 308 K)

Final temperature = 55°C (55+273 = 328 K)

Final volume = ?

Final pressure = 4 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 2 atm ×200 mL × 328 K / 308 K ×4 atm

V₂ = 131200 atm .mL. K / 1232 K.atm

V₂ = 106.5 mL

6 0
2 years ago
A 2-kg block rests on a wall that is 3 m tall. What is the gravitational potential energy of the block? The force of gravity on
Keith_Richards [23]
PE= weight x height. (remember weight is the force of gravity acting on objects...NOT the same as mass)

PE=3*2*9.8
PE=6*9.8
PE=58.8 J (for joules which is the unit for both Kinetic and potential energy)

Hope this helps :D
7 0
3 years ago
A compound distributes between benzene (solvent 1) and water (solvent 2) with a distribution coefficient, K = 2.7. If 1.0g of th
mote1985 [20]

Explanation:

The given data is as follows.

Solvent 1 = benzene,          Solvent 2 = water

 K_{p} = 2.7,         V_{S_{2}} = 100 mL

V_{S_{1}} = 10 mL,       weight of compound = 1 g

       Extract = 3

Therefore, calculate the fraction remaining as follows.

                  f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}

                                  = [1 + 2.7(\frac{100}{10})]^{-3}

                                  = (28)^{-3}

                                  = 4.55 \times 10^{-5}

Hence, weight of compound to be extracted = weight of compound - fraction remaining

                                  = 1 - 4.55 \times 10^{-5}

                                  = 0.00001

or,                               = 1 \times 10^{-5}

Thus, we can conclude that weight of compound that could be extracted is 1 \times 10^{-5}.

7 0
3 years ago
Read 2 more answers
Calculate the number of moles in a 14.5 gram sample of C4H10.
cricket20 [7]
Moles= mass divided by molar mass
Molar mass= 12.01(4) + 1.01(10)
                 = 58.14g/mol

Moles=14.5g / 58.14g/mol
         =0.249

Therefore there are approx 0.249 moles in a 14.5g sample of C4H10
6 0
3 years ago
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