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Amanda [17]
3 years ago
6

I'm afraid I won't pass. And very nervous. Can you help me please.

Chemistry
1 answer:
daser333 [38]3 years ago
4 0
Just look at the number in front also called coefficient (you have to balance the equations first, but all the questions here are balanced, so no worries). for q1.
in the balanced equation, the number in front of aluminum oxide is 2 (2 - this number Al2O3) and for aluminium is 4 as in (4 Al). so the ratio is 2:4. simplified it is 1:2. or write it out fully
2 Al2O3: 4 Al
ignore everything after the number.
2:4
same as 1:2
Aluminium oxide to oxygen
2 Al2O3: 3 O2
2:3
aluminum to oxygen
4 Al: 3 O2
4:3

question 2
Mercury oxide to Mercury
2 HgO : 2 Hg
2:2
same as 1:1
Mercury oxide to oxygen
2 HgO : O2
since oxygen in this case does not have a number written in front of it, the default is 1.
2: 1.
you should be able to do the rest
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3 years ago
Who no this i need help i need to pass to the next grade
Hitman42 [59]

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3 years ago
Balanced nuclear equation forces the alpha decay Po-210
Serjik [45]

Answer:

_{84}^{210}\text{Po} \longrightarrow \, _{82}^{206}\text{Pb} \, + \, _{2}^{4}\text{He}

Explanation:

Your unbalanced nuclear equation is:

_{84}^{210}\text{Po} \longrightarrow \, _{x}^{y}\text{Z} + \, _{2}^{4}\text{He}

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.  

Then

 84 = x + 2, so x =  84 - 2 =   82

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_{84}^{210}\text{Po} \longrightarrow \, _{82}^{206}\text{Pb} \, + \, _{2}^{4}\text{He}

8 0
3 years ago
A lead mass is heated and placed in a foam cup calorimeter containing 40.0 mL of water at 17.0°C. The water reaches a temperatur
lbvjy [14]

Answer: 502 Joules

Explanation:

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 40.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{40.0mL}\\\\\text{Mass of water}=(1g/mL\times 40.0mL)=40.0g

When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

q=m\times c\times \Delta T

q = heat absorbed by water

m = mass of water = 40.0 g

T_{final} = final temperature of water = 20.0°C

T_{initial = initial temperature of water = 17.0°C

c = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

q=40.0\times 4.186\times (20.0-17.0)]

q=502J

Hence, the joules of heat were re-leased by the lead is 502

5 0
3 years ago
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Answer:

Explanation:

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