Use your calculator to determine the answer to the following calculation:<br> 1.0×10^−15/4.2×10^−7

What kind of bond holds together atoms within a molecule?

sesenic [268]

I think it would be these three answers ionic , covalent , and polar covalent

7 0

3 years ago

Read 2 more answers

What is the potential energy of the bowling ball as it sits on top of the building?

Tju [1.3M]

Answer:

Potential energy is energy due to an object's height above the ground.

Potential energy = mass x gravity x height

Kinetic energy is energy due to the motion of the object.

Kinetic energy = 1/2 x mass x velocity²

8 0

3 years ago

When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H

Taya2010 [7]

Answer:

- Empirical:

$C_3H_7$

- Molecular:

$C_6H_{14}$

Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

$n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC$

Moreover, the moles of hydrogen:

$n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH$

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

$C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}$

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

$C_3H_7$

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

$C_6H_{14}$

Which is hexane.

Best regards.

6 0

3 years ago

Carry out the following operations as if they were calculations of experimental results and express each answer in standard nota

Shalnov [3]

Answer:

1. $10.6\; \rm m$ (one decimal place.)

2. $0.79\; \rm g$ (two decimal places.)

3. $16.5\;\rm cm^2$ (three significant figures.)

Explanation:

The first and second expressions are additions and subtractions. When adding two numbers, the accuracy of the result is given by the number of decimal places in it. The result should have as many decimal places as the input with the least number of decimal places.

For example, in the first expression:

$5.6792\;\rm m$ has four decimal places.
$0.6\; \rm m$ has only one decimal place.
$4.33\; \rm m$ has two decimal places.

Therefore, the result should be rounded to one decimal place. Note that these units are compatible for addition, since they are all the same. The result should have the same unit (that is: $\rm m$ .)

Therefore:

$\rm 5.6792\; m + 0.6\; m + 4.33\; m \approx 10.6\; \rm m$ . (Rounded to one decimal place.)

Similarly:

$\rm 3.70\; \rm g$ has two decimal places.
$2.9133\; \rm g$ has four decimal places.

Therefore, the result should be rounded to two decimal places. Its unit should be $\rm g$ (same as the unit of the two inputs.)

$\rm 3.70\; g - 2.9133\; g \approx 0.79\; \rm g$ . (Rounded to two decimal places.)

When multiplying two numbers, the accuracy of the result should be based on the number of significant figures in it. The result should have as many significant figures as the input with the least number of significant figures. In this expression:

$4.51\; \rm cm$ has three significant figures.
$3.6666\; \rm cm$ has five significant figures.

Therefore, the result should have only three significant figures.

The unit of the result is supposed to be the product of the units of the input. In this expression, that unit will be $\rm cm \cdot cm$ , which is occasionally written as $\rm cm^2$ .