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BlackZzzverrR [31]
3 years ago
6

Can a physical or chemical change the amount of matter present

Chemistry
1 answer:
dezoksy [38]3 years ago
5 0

Extensive properties, such as mass and volume, depend on the amount of matter being measured. Intensive properties, such as density and color, do not depend on the amount of the substance present. Physical properties can be measured without changing a substance's chemical identity.

You might be interested in
Write 0.00103 in scientific notation
Andrei [34K]

Answer:

1.03

Explanation:

You would take 0.00103 and move the decimal like this; 0001.03, we wouldn't have the zeroes in front of the one, as it can throw us off. Therefore, your answer would be 1.03.

Hope I helped! Don't hesitate to let me know if I made a mistake.

3 0
3 years ago
(will give brainliest) show your work. How many grams of Copper(I) nitrate, CuNO3 are required to produce 88.0 grams of aluminum
ValentinkaMS [17]

Based on the stoichiometry of the reaction, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

<h3>What is stoichiometry of a reaction?</h3>

The stoichiometry of a reaction is the molar ratio in which reactants combine to form products.

The stoichiometry of the reaction shows that 6 moles of copper (i) nitrate produces 2 moles of aluminium nitrate.

molar mass of Copper(I) nitrate, CuNO3 = 126 g

molar mass of aluminum nitrate, Al(NO3)3 = 213 g

88.0 g of aluminum nitrate, Al(NO3)3 = 88.0/213 moles = 0.413 moles

0.413 moles of Al(NO3)3 will be produced by 0.413 ×6/3 = 1.239 moles of CuNO3

Mass of 1.239 moles of CuNO3 = 1.239 × 126 = 156.114 g of CuNO3

Therefore, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

Learn more about stoichiometry at: brainly.com/question/16060223

Therefore, 156.114 g of CuNO3

4 0
2 years ago
Hydrazine (N2H4), a rocket fuel , reacts with oxygen to form nitrogen gas and water vapor. The reaction is represented with the
lilavasa [31]

Answer:

D)  8.40 L H₂O(g).

Explanation:

  • The balanced equation for the mentioned reaction is:

<em>2N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(g), </em>

It is clear that 2.0 moles of N₂H₄ react with 1.0 mole of O₂ to produce 2.0 moles of N₂ and 2.0 moles of H₂O.

  • At STP, 4.20L of O₂ reacts with N₂H₄:

It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.

<u>using cross multiplication: </u>

1.0 mol of O₂ represents → 22.4 L.

??? mol of O₂ represents → 4.2 L.

∴ 4.2 L of O₂ represents = (1.0 mol)(4.2 L)/(22.4 L) = 0.1875 mol.

  • To find the no. of moles of H₂O produced:

Using cross multiplication:

1.0 mol of O₂ produce → 2.0 mol of H₂O, from stichiometry.

0.1875 mol of O₂ produce → ??? mol of H₂O.

∴ The no. of moles of H₂O = (2.0 mol)(0.1875 mol)/(1.0 mol) = 3.75 mol.

  • Again, using cross multiplication:

1.0 mol of H₂O represents → 22.4 L, at STP.

3.75 mol of H₂O represents → ??? L.

<em>∴ The no. of liters of water vapor will be produced </em>= (3.75 mol)(22.4 L)/(1.0 mol) = <em>8.4 L.</em>

<em></em>

<em>So, the right choice is: D)  8.40 L H₂O(g).</em>

<em></em>

3 0
3 years ago
20 POINTS!!
Elza [17]
The answer is option number one
4 0
3 years ago
Read 2 more answers
Solution A is 0.44 M and reacts with 0.11 M of solution B. Assume that the value of x is 0, the value of y is 1, and r is 1.07 ×
xz_007 [3.2K]

Answer:

K, the rate constant = 9.73 × 10^(-1)/s

Explanation:

r = K × [A]^x × [B]^y

r = Rate = 1.07 × 10^(-1)/s

K = Rate constant

A and B = Concentration in mol/dm^-3

A = 0.44M

B = 0.11M

x = Order of reaction with respect to A = 0

y = Order of reaction with respect to B = 1

Solving, we get

r/([A]^x × [B]^y) = K

K = 1.07 × 10^(-1)/s/(0.44^0 × 0.11^1)= 0.9727

K = 0.9727

7 0
3 years ago
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