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3 years ago

A block weighing 13.15 grams has sides of 11.5 cm, 5.75 cm, and 8.44 cm. What is the density of the block in g/cm?

Chemistry

1 answer:

vitfil [10]3 years ago

7 0

Answer:

B. or .0236

Explanation:

Multiply the 3 sides to get 558.095, then divide the mass and volume to get .0236

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AysviL [449]

Answer: pH = 14

Explanation: Please see the attachments below

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3 years ago

15.630 g milk chocolate bar ks found to contain 9.815g of sugar. how many milligrams of sugar does the bar contain?

sleet_krkn [62]

Well there are 1,000 milligrams in 1 gram so 9.815 grams of sugar is equivalent to 9,815 milligrams of sugar. So a bar of milk chocolate contains 9,815 milligrams of sugar.

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3 years ago

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A gas starts at a volume of 23 L and a pressure of 1.23 atm. What is the new pressure if you

Anna71 [15]

Answer:

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

From the question we have

$V_2 = \frac{1.23 \times 23}{15} = \frac{28.29}{15} \\ = 1.886$

We have the final answer as

Hope this helps you

7 0

3 years ago

Ham radio operators often broadcast on the 6-meter band. The frequency of this electromagnetic radiation is MHz. [speed of light

Volgvan

Answer:

50 MHz

Explanation:

The relation between frequency and wavelength is shown below as:

$c=frequency\times Wavelength$

c is the speed of light having value $3\times 10^8\ m/s$

Given, Wavelength = 6 m

Thus, Frequency is:

$Frequency=\frac{c}{Wavelength}$

$Frequency=\frac{3\times 10^8}{6}\ Hz$

$Frequency=50\times 10^{6}\ Hz$

Also,

Also, 1 Hz = $10^{-6}$ MHz

So,<u> Frequency = 50 MHz.</u>

8 0

3 years ago

A standard 10.00 g mass is weighed on an analytical balance 100 times. The average and standard deviation obtained gives 10.12 ±

Mekhanik [1.2K]

Answer:

There was an improvement in accuracy. There was no change in precision.

Explanation:

<em>The average mass after recalibration is closer to the mass of the standard, </em>so the recalibration improved the accuracy<em> </em>(the measurement is closer to an accepted 'true' value).

The standard deviation did not change, so the precision (or how disperse the measurements are) was not affected.

5 0

3 years ago