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Zinaida [17]
3 years ago
12

The amount of matter in an object is its ____ weight mass volume density

Chemistry
1 answer:
Ahat [919]3 years ago
5 0
The amount of mater in a object is called matter
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Elina [12.6K]

H2SO4 + ZN ------- ZNSO4+ H2

(SO4)²The sulphate salt is formed......

Hope it helps

8 0
3 years ago
Write a net ionic equation to show that acetylsalicylic acid (aspirin), hc9h7o4, behaves as a brønsted-lowry acid in water.
Thepotemich [5.8K]
By definition, Bronsted-Lowry acid is a proton donor in the acid-base neutralization reaction. When a weak acid like acetylsalicylic acid is reacted with water, the water here acts as the Bronsted-Lowry base. This is possible because water has properties of amphoterism - can act as an acid or base. In this case, acetylsalicylic acid would have to donate its H+ atom to water, so that it would yield a hydronium ion, H₃O⁺. The complete net ionic reaction is shown in the picture.

So, in the reaction, the products yield are the acetylsalicylate ion and the hydronium ion.

3 0
3 years ago
Are there any plans for controlling or removing zebra mussel
Sav [38]

Answer: no

Explanation:

4 0
3 years ago
Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow
andreyandreev [35.5K]

Explanation:

(a)  The given data is as follows.

              Pressure on top (P_{o}) = 140 bar = 1.4 \times 10^{7} Pa       (as 1 bar = 10^{5})

              Temperature = 15^{o}C = (15 + 273) K = 288 K

         Density of gas = \frac{PM}{ZRT}

                \frac{dP}{dZ} = \rho \times g

               \frac{dP}{dZ} = \int \frac{PM}{ZRT}

                \int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ

           ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}

                              = 0.4548

                     P_{1} = P_{o} \times e^{0.4548}

                                 = 1.4 \times 10^{7} Pa \times 1.5797

                                 = 2.206 \times 10^{7} Pa

Hence, pressure at the natural gas-oil interface is 2.206 \times 10^{7} Pa.

(b)   At the bottom of the tank,

                 P_{2} = P_{1}  + \rho \times g \times h

                             = 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

                             = 309.8 \times 10^{5} Pa

                             = 309.8 bar

Hence, at the bottom of the well at 15^{o}C pressure is 309.8 bar.

6 0
4 years ago
Which of the following are examples of types of mixtures?
aev [14]
I think it would be homogeneous and heterogeneous
6 0
3 years ago
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