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2 years ago

14

Fossils can be body parts of ancient organisms, or they can be traces. Give five examples of traces. Help this is due today!!!!!

7 grade

1 answer:

SSSSS [86.1K]2 years ago

6 0

Burrows, tracks, coprolites, nests and footprints are examples of traces that can be found in a fossil.

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What is the oxidation number for Zn(OH)4 2-?

AleksAgata [21]

<span>Zn(OH)4= ; -2 = -4 + ?, Zn = +2 , the -4 is the charge on the 4 OH- ions.

NO3- + 6 H2O + 8 e- -------> NH3 + 9 OH-
( Zn + 4 OH- ----> Zh(OH)4= + 2e-) x 4
7 OH- + NO3- + 4 Zn + 6 H2O -------> NH3 + 4 Zn(OH)4</span>

8 0

2 years ago

When a candle burns it produced 41,300 Joules per 1 gram. Use dimensional analysis to convert this to Calories per pound.

insens350 [35]

Answer:

4477381.7 calories/pound

Explanation:

It is given that,

When a candle burns it produced 41,300 Joules per 1 gram.

We need to convert it into calories per pound.

We know that,

1 cal = 4.184 J

⇒ 1 J = (1/4.184) cal

1 pound = 453.592 grams

⇒1 g = (1/453.592) pounds

Now,

$41300\ \text{Joules per gram}=\dfrac{41300\ \text{Joules}}{\text{gram}}\\\\=41300\times \dfrac{\dfrac{1}{4.184}\ \text{calories}}{\dfrac{1}{453.592}\ \text{gram}}\\\\=4477381.7\ \text{calories/pound}$

Hence, 41,300 Joules/gram = 4477381.7 calories/pound.

3 0

2 years ago

Which is a component of John Dalton’s atomic theory?

s344n2d4d5 [400]

I believe its a...................

7 0

2 years ago

Read 2 more answers

3. Which is not true about natural<br> selection?

tatuchka [14]

Answer:

A:There must be diversity in the population.

Explanation:

Hopefully this helps!

5 0

2 years ago

Suppose 550.mmol of electrons must be transported from one side of an electrochemical cell to another in 49.0 minutes. Calculate

nekit [7.7K]

Answer:

18.0 Ampere is the size of electric current that must flow.

Explanation:

Moles of electron , n = 550 mmol = 0.550 mol

1 mmol = 0.001 mol

Number of electrons = N

$N=N_A\times n$

Charge on N electrons : Q

$Q = N\times 1.602\times 10^{-19} C$

Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds

1 min = 60 seconds

Size of current : I

$I=\frac{Q}{T}=\frac{N\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}$

$=\frac{n\times N_A\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}$

$I=\frac{0.550 mol\times 6.022\times 10^{23} mol^{-1}\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}=18.047 A\approx 18.0 A$

18.0 Ampere is the size of electric current that must flow.

3 0

2 years ago