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3 years ago

What can we use to determine the relative age of two rocks

1 answer:

7 0

You can use the law of superposition:in horizontal sedimentary rock,the oldest layer is at the bottom.the more layers up is younger than the ones under

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A 100g sample of Carbon-14 has a half-life of 5 years. How much Carbon-14 is left after 10 years?

attashe74 [19]

Answer: The amount of carbon-14 left after 10 years is 25 g

Explanation:

Formula used :

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives = ?

$a_o$ = Initial amount of the reactant = 100 g

n = number of half lives = $\frac{\text {given time}}{\text {half life}}=\frac{10}{5}=2$

Putting values in above equation, we get:

$a=\frac{100g}{2^2}$

$a=\frac{100g}{4}=25g$

Therefore, the amount of carbon-14 left after 10 years is 25 g

6 0

3 years ago

How many grams of KCIO3 are needed to produce 5.00 of O2 at STP?

Arlecino [84]

2KClO₃ → 2KCl + 3O₂

mole ratio of KClO₃ to O₂ is 2 : 3

∴ if moles of O₂ = 5 mol

then moles of KClO₃ = $\frac{5 mol * 2}{3}$

= 3.33 mol

Mass of KClO₃ needed = mol of KClO₃ × molar mass of KClO₃

= 3.33 mol × ((39 × 1) + (35.5 × 1) + (16 × 3) g/mol

= 407.93 g

6 0

2 years ago

6) Which of these elements is a metal? E) C D) N A) O C) Ne B) Ca

sertanlavr [38]

Answer:

Explanation:

It is on the left side of the periodic table where metals are located.

4 0

3 years ago

Read 2 more answers

When solving a problem it is important to identify your given and needed units, but it is also important to understand the relat

UNO [17]

Answer:

See Explanation

Explanation:

Given

(a) to (d)

Required

Determine whether the given parameters can calculate the required parameter

To calculate either Density, Mass or Volume, we have

$Density = \frac{Mass}{Volume}$

$Mass = Density * Volume$

$Volume = \frac{Mass}{Density}$

(a) 432 g of table salt occupies 20.0 cm^3 of space

Here, we have:

$Mass = 432g$

$Volume = 20.0cm^3$

The above can be used to calculate Density as follows;

$Density = \frac{Mass}{Volume}$

$Density = \frac{432g}{20.0cm^3}$

$Density = 21.6g/cm^3$

(b) 5.00 g of balsa wood, density of balsa wood : 0.16 g/cm^3

Here, we have:

$Mass = 5.00g$

$Density = 0.16g/cm^3$

This can be used to solve for Volume as follows:

$Volume = \frac{Mass}{Density}$

$Volume = \frac{5.00g}{0.16g/cm^3}$

$Volume = 31.25cm^3$

(c) 32 cm^3 sample of gold density of 19.3 g/cm^3

Here, we have:

$Volume = 32cm^3$

$Density = 19.3g/cm^3$

This can be used to calculate Mass as follows:

$Mass = Density * Volume$

$Mass = 32cm^3 * 19.3g/cm^3$

$Mass = 617.6g$

(d) 150 g of iron, density of Iron = 79.0 g/cm^3

Here, we have

$Mass = 150g$

$Density = 79.0g/cm^3$

This can be used to calculate volume as follows:

$Volume = \frac{Mass}{Density}$

$Volume = \frac{150g}{79.0g/cm^3}$

$Volume = 1.90 cm^3$ Approximated

5 0

3 years ago

What is special about the kelvin temperature scale

nalin [4]

Kelvin is a temperature scale designed so that zero degrees K is defined as absolute zero (at absolute zero, a hypothetical temperature, all molecular movement stops - all actual temperatures are above absolute zero) and the size of one unit is the same as the size of one degree Celsius.

5 0

3 years ago