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3 years ago

Keeping in mind the rules for significant digits, what is the product of 0.8954 m and 2.98 m?

Chemistry

1 answer:

Shalnov [3]3 years ago

3 0

Answer: 3.240576

Explanation:

There are various ways of doing multiplication this is one of them

The wording "keep in mind the rules for significant digits" implies that we are to use these to show where the decimal place goes by approximating the answer. Then conducting the exact calculation.

Consider 0.8 as 8

2.4

So the solution is some where a bit above 2.4

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Lets just use the numbers

Multiply both 0.8439 and 3.84 by 1000 to get rid of the decimal place. This makes thing more straight forward. So we end up with:

8439

38400

Split up the 38400 into 30000+ 8000 + 400

30000

8439

253170000

8000

8439

67512000

400

8439

3375600

−−−−−−−−−

←

Add

324057600

By our approximation we are expecting the form of

2.4

small amount

So the answer is

3.240576

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Check by calculator:

0.8439

3.84

3.240576

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What is the final temperature of a sample of water if 12.15J of energy is added to 475g of water at an initial temperature of 20

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Answer:

The final temperature of water is 20.5061 °C.

Explanation:

Let the final temperature of water be 'x'.

Given:

Heat added to water is, $Q_{in}=12.15\ J$

Initial temperature of water is, $T=20.5\ \°C$

Mass of water is, $m=475\ g$

Now, heat is added to water and its temperature is increased. The temperature is increased because water absorbs all the heat.

Heat absorbed by water is given as:

$Q_{abs}=mc(x-T)$ where 'c' is specific heat capacity of water and its value is equal to 4.186 J/g °C.

Now, plug in the given values and simplify.

$Q_{abs}=475\times4.186\times(x-20.5)\\Q_{abs}=1988.35(x-20.5)$

Now, from law of conservation of energy, we know that:

Heat absorbed by water = Heat added to water

$Q_{abs}=Q_{in}\\1988.35(x-20.5)=12.15\\x-20.5=\frac{12.15}{1988.35}\\x-20.5=0.0061\\x=20.5+0.0061=20.5061\ \°C$

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When 50.0 mL of 1.00 M HCl(aq) and 50.0 mL of 1.00 M NaOH(aq) , both at 22.0 °C, are added to a coffee cup calorimeter, the temp

Igoryamba

Answer:

The heat produced by this reaction is 2887J

Explanation:

The reaction of HCl with NaOH occurs as follows:

HCl + NaOH → H2O + NaCl + ΔH

Where ΔH is released heat due the reaction

The equation of a coffee cup calorimeter is:

Q = C*m*ΔT

Where Q is heat produced,

C is specific heat of the solution (4.18J/g°C),

m is mass of the solution (50mL + 50mL = 100mL = 100g -Because density is 1g/mL)

and ΔT is change in temperature (28.9°C - 22.0°C = 6.9°C)

Replacing:

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3 years ago

Keeping in mind the rules for significant digits, what is the product of 0.8954 m and 2.98 m?​

Keeping in mind the rules for significant digits, what is the product of 0.8954 m and 2.98 m?