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3 years ago

Give the result of the following mathematical operations to the correct number of significant figures. 5.659 * (3.496 - 2.814) =

Chemistry

1 answer:

shtirl [24]3 years ago

7 0

Answer:

The result is 3.859 in which we use four significant figures.

Explanation:

We start by solving the mathematical operation :

$5.659.(3.496-2.814)=5.659(0.682)=3.859438$

The result for the operation is 3.859438 but the numbers in the operation are given with four significant figures and that is why we are going to use four significant figures to express the result

To express 3.859438 with four significant figures we use the first four digits that appear from left to right starting by the first digit that is different to zero

In this case : 3.859 will be the result with four significant figures.

We also use a rule that says : To decide if the last significant figure remains the same we look for the value of the digit at its right.

If that number is greater than or equal to 5 ⇒ we sum one to the last significant figure

For example 3.859738 = 3.860 with four significant figures because the ''7'' is greater that 5

If that number is less than 5 ⇒ the last significant figure remains the same

In our case : 3.859438 = 3.859 because ''4'' is less than ''5''

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What mass of solid that has a molar mass of 46.0 g/mol should be added to 150.0 g of benzene to raise the boiling point of benze

enyata [817]

Answer : 17.12 g

Explanation: $\Delta T =k_b\times m$

$\Delta T$ = elevation in boiling point

$k_b$ = boiling point elevation constant

m= molality

$molality=\frac{mass of solute}{molecular mass of solute\times weight of the solvent in kg}$

given $\Delta T=6.28^{\circ}C$

Molar mass of solute = 46.0 $gmol^{-1}$

Weight of the solvent = 150.0 g = 0.15 kg

Putting in the values

$molality=\frac{x}{46gmol^{-1}\times0.15kg}$

$6.28 =2.53^{\circ}Ckgmol^{-1}\frac{x}{46gmol^{-1}\times 0.15kg}$

x = 17.12 g

3 0

3 years ago

If the same amount of heat is added to 25.0 g of each of the metals, which are all at the same initial temperature, which metal

AVprozaik [17]

Answer:

The bismuth sample.

Explanation:

The specific heat $c$ of a substance (might not be a metal) is the amount of heat required for heating a unit mass of this substance by unit temperature (e.g., $\rm 1\; ^{\circ}C$ .) The formula for specific heat is:

$\displaystyle c = \frac{Q}{m \cdot \Delta T}$ ,

where

$Q$ is the amount of heat supplied.
$m$ is the mass of the sample.
$\Delta T$ is the increase in temperature.

In this question, the value of $Q$ (amount of heat supplied to the metal) and $m$ (mass of the metal sample) are the same for all four metals. To find $\Delta T$ (change in temperature,) rearrange the equation:

$\displaystyle c \cdot \Delta T = \frac{Q}{m}$ ,

$\displaystyle \Delta T = \frac{Q}{c \cdot m}$ .

In other words, the change in temperature of the sample, $\Delta T$ can be expressed as a fraction. Additionally, the specific heat of sample, $c$ , is in the denominator of that fraction. Hence, the value of the fraction would be the largest for sample with the smallest specific heat.

Make sure that all the specific heat values are in the same unit. Find the one with the smallest specific heat: bismuth ( $\rm 0.123 \; J \cdot g\cdot \,^{\circ}C^{-1}$ .) That sample would have the greatest increase in temperature. Since all six samples started at the same temperature, the bismuth sample would also have the highest final temperature.

3 0

3 years ago

stbank, Question 075 Get help answering Molecular Drawing questions. Compound A, C6H12 reacts with HBr/ROOR to give compound B,

Law Incorporation [45]

Answer:

Explanation:

In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).

A and C reacts with two differents reagents and conditions, however both of them gives the same product.

Let's analyze each reaction.

First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.

Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.