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zaharov [31]
2 years ago
7

you have two samples of gray powder both which are flammable these are powders the same substance explain why ​

Chemistry
2 answers:
nasty-shy [4]2 years ago
7 0

The amount of matter stays the same between the substances,

Jet001 [13]2 years ago
5 0

Answer:it help in gravity

Explanation:

You might be interested in
Combustion of 8.652 g of a compound containing c, h, o, and n yields 11.088 g of coz, 3.780 g of h2o, and 3.864 g of no2. How ma
Katyanochek1 [597]

Answer:-  C = 3.024 g, H = 0.42 g, N = 1.176 g and O = 4.032 g

Solution:- The compound contains C, H, N and O. On combustion, all the carbon is converted to carbon dioxide, All hydrogen is converted to water and all the nitrogen is converted to nitrogen dioxide.

From the grams of all these, we could calculate their moles and then using mol ratio of these products and the number of moles of C, H and N present in them, we calculate the moles of C, H and N respectively. Further, these moles are converted to grams. On subtracting the sum of grams of C, H and N from the mass of the sample, the mass of oxygen is calculated.

The calculations are as follows:

Calculations for the grams of C:-

11.088gCO_2(\frac{1molCO_2}{44gCO_2})(\frac{1molC}{1molCO_2})(\frac{12gC}{1molC})

= 3.024 g C

Calculations for the grams of H:-

3.780gH_2O(\frac{1molH_2O}{18gH_2O})(\frac{2molH}{1molH_2O})(\frac{1gH}{1molH})

= 0.42 g H

Calculations for the grams of N:-

3.864gNO_2(\frac{1molNO_2}{46gNO_2})(\frac{1molN}{1molNO_2})(\frac{14gN}{1molN})

= 1.176 g N

Mass of the compound is given as 8.652 g. Now we could calculate the grams of oxygen as:

mass of oxygen = 8.652 - (3.024 + 0.42 + 1.176)

= 8.652 - 4.62

= 4.032 g

So, 8.652 grams of the compound contains 3.024 g of C, 0.42 g of H, 1.176 g of N and 4.032 g of O.


7 0
1 year ago
2C6H5COOH + 15O2 → 14CO2 + 6H2O which of the following options gives the correct product:product ratio?
mash [69]
You did not include the options but I can tell you the product ratio.

The product ratio is the mole ratio of the products of the reaction.

From the balanced chemical equation you have all the mole ratios:

The given equation is: 2 C6H5COOH + 15O2 --> 14 CO2 + 6H2O

The mole ratios are: 2 C6H5COOH: 15 O2: 14 CO2 : 6 H2O

The products are CO2 and H2O

Their mole ratio = 14 CO2 : 6 H2O

That can be expressed as:

14 mol CO2        7 mol CO2
----------------- =  -----------------
  6 mol H2O        3 mol H2O

It is also the same that:

6 mol H2O : 14 mol CO2

  6 mol H2O           3 mol H2O
------------------ =  -------------------
14 mol CO2           7 mol CO2

So, compare your options to the ratios show above and pick the proper ratio.
7 0
2 years ago
Isoniazid is used in the treatment of tuberculosis and multiple sclerosis. Identify each lone pair as either localized or deloca
Sholpan [36]

Answer:

Please find the complete question and its solution in the attached file:

Explanation:

3 0
1 year ago
At 15°C, the value of Kw is 4.5 × 10⁻¹⁵. What is the equilibrium concentration of OH⁻ at this temperature?
MaRussiya [10]
[H3O+][OH-] = 4.5 x 10^-15
[H3O+] = [OH-] = 6.7 x 10^-8 M

Explanation: do the math
5 0
2 years ago
A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro
Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on  combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q

Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J

Heat released on  combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows  :

Heat released on  combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q'

Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J

Heat released on  combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0
1 year ago
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