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Ipatiy [6.2K]
2 years ago
10

If Zn and H2SO4 undergo a single-displacement reaction, what is the balanced equation?

Chemistry
1 answer:
ankoles [38]2 years ago
6 0

Answer:Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

O2Zn(s) + H2SO4(aq) → 22nH(aq) + SO4(s)

Explanation:

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Answer:

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4 0
2 years ago
Which of the following is an example of using creativity while recording measurements during an experiment?
Oksanka [162]

I’d have to go with B

5 0
3 years ago
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Approximately how many ice cubes must melt to cool 650 milliliters of water from 29°C to 0°C? Assume that each ice cube contains
qwelly [4]

Answer : The number of ice cubes melt must be, 13

Explanation :

First we have to calculate the mass of water.

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}

Density of water = 1.00 g/mL

Volume of water = 650 mL

\text{Mass of water}=1.00g/mL\times 650mL=650g

Now we have to calculate the heat released on cooling.

Heat released on cooling = m\times c\times (T_2-T_1)

where,

m = mass of water = 650 g

c = specific heat capacity of water = 4.18J/g^oC

T_2 = final temperature = 29^oC

T_2 = initial temperature = 0^oC

Now put all the given values in the above expression, we get:

Heat released on cooling = 650g\times 4.18J/g^oC\times (29-0)^oC

Heat released on cooling = 78793 J = 78.793 kJ   (1 J = 0.001 kJ)

As, 1 ice cube contains 1 mole of water.

The heat required for 1 ice cube to melt = 6.02 kJ

Now we have to calculate the number of ice cubes melted.

Number of ice cubes melted = \frac{\text{Total heat}}{\text{Heat for 1 ice cube}}

Number of ice cubes melted = \frac{78.793kJ}{6.02kJ}

Number of ice cubes melted = 13.1 ≈ 13

Therefore, the number of ice cubes melt must be, 13

3 0
3 years ago
Write the symbol for every chemical element that has atomic number greater than 55 and atomic mass less than 144.0 u
laila [671]
You will need a periodic table to help you answer this problem. The atomic numbers are arrange from lowest to highest in the periodic table. You can locate element number 55 to be Cesium with an atomic weight of 132.905 amu. So, you start from element 56. The following elements are:

56     Barium       137.328 amu
57     Lanthanium   138.905 amu
58     Cerium       140.116 amu
59     <span>Praseodymium    140.908 amu
60     Neodymium   144.243 amu

Neodymium is already greater than 144 amu. Therefore, these elements only include Barium, Lanthanium, Cerium and Praseodymium.</span>
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3 years ago
How did your estimated poputation size<br> compare with the actual population size
musickatia [10]

Answer:

Hope this helps

Explanation:

https://seagrant.whoi.edu/wp-content/uploads/2018/05/ESTIMATING-POPULATION-SIZE-1.pdf

8 0
3 years ago
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