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3 years ago

If it takes 2000 years to weather 1 cm of limestone, how long will it take to weather 5 cm of limestone?

Chemistry

1 answer:

Anna71 [15]3 years ago

3 0

Answer:10,0000 years

Explanation:

If it takes 2000 years to weather 1cm of limestone

It will take 5*2000 years to weather 5cm of limestone. This gives us 10000 years.

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If I have 6 moles of a gas at a pressure of 3.4 atm and a volume of 10 liters, what

ch4aika [34]

Answer:

The temperature is 69.05 K.

Explanation:

We have,

Number of moles are 6

Pressure is 3.4 atm and a volume of 10 liters.

It is required to find the temperature. It can be calculated using gas law equation. It says that,

$PV=nRT$

P is pressure

V is volume

n is number of moles

R is gas constant, $R=0.082057\ L-atm/mol-K$

T is temperature

Plugging all the values we get :

$T=\dfrac{PV}{nR}\\\\T=\dfrac{3.4\times 10}{6\times 0.082057}\\\\T=69.05\ K$

So, the temperature is 69.05 K.

8 0

3 years ago

Why does it seem to us as if rocks remain the same and don't change

qaws [65]

Most rocks that we encounter in our normal everyday lives are sedimentary rocks. Sedimentary rocks are rocks that have been worn down gradually over long periods of time. Because it takes very long periods of time (couple decades) for these rocks to change, it often seems as if they don't change at all, when in reality the change is too small for us to realize it!

3 0

3 years ago

A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm

Ulleksa [173]

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100$

$Percent error = \frac{0.1}{63.2}\times 100$

$Percent error = 0.00158\times 100$

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100$

$Percent error = \frac{0.2}{63.2}\times 100$

$Percent error = 0.00316\times 100$

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100$

$Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100$

$Percent error = \frac{(0.5)}{63.2}\times 100$

$Percent error = 0.00791\times 100$

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%

7 0

3 years ago

please help i dont get chemistry as much as I've only just started and I dont know how to separate all these matters

dsp73

is this technical chemistry?

6 0

2 years ago

A gold coin contains 3.47 × 10^23<br> gold atoms.<br> What is the mass of the coin in grams?

VashaNatasha [74]

Answer:

Mass = 114.26 g

Explanation:

Given data:

Number of gold atoms = 3.47×10²³ atoms

Mass in gram = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

3.47×10²³ atoms × 1 mol /6.022 × 10²³ atoms

0.58 mol

Mass of gold:

Mass = number of moles × molar mass

Mass = 0.58 mol × 197 g/mol

Mass = 114.26 g

4 0

3 years ago