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3 years ago

Which of these civilizations still has strong cultural and religious ties to their star

Chemistry

1 answer:

aliya0001 [1]3 years ago

6 0

Answer:

In the 2nd millennium, the eastern coastlines of the Mediterranean are dominated by the Hittite and Egyptian empires, competing for control over the city states in the Levant (Canaan)

Explanation:

because it is

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Combustion analysis of toluene, a common organic solvent, gives 3.52 mg of co2 and 0.822 mg of h2o. if the compound contains onl

IRISSAK [1]

<span>C7H8 First, lookup the atomic weight of all involved elements Atomic weight of carbon = 12.0107 Atomic weight of hydrogen = 1.00794 Atomic weight of oxygen = 15.999 Then calculate the molar masses of CO2 and H2O Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol Now calculate the number of moles of each product obtained Note: Not interested in the absolute number of moles, just the relative ratios. So not going to get pedantic about the masses involved being mg and converting them to grams. As long as I'm using the same magnitude units in the same places for the calculations, I'm OK. moles CO2 = 3.52 / 44.0087 = 0.079984 moles H2O = 0.822 / 18.01488 = 0.045629 Since each CO2 molecule has 1 carbon atom, I can use the same number for the relative moles of carbon. However, since each H2O molecule has 2 hydrogen atoms, I need to double that number to get the relative number of moles for hydrogen. moles C = 0.079984 moles H = 0.045629 * 2 = 0.091258 So we have a ratio of 0.079984 : 0.091258 for carbon and hydrogen. We need to convert that to a ratio of small integers. First divide both numbers by 0.079984 (selected since it's the smallest), getting 1: 1.140953 The 1 for carbon looks good. But the 1.140953 for hydrogen isn't close to an integer. So let's multiply the ratio by 1, 2, 3, 4, ..., etc and see what each new ratio looks like (Effectively seeing what 1, 2, 3, 4, etc carbons look like) 1 ( 1 : 1.140953) = 1 : 1.140953 2 ( 1 : 1.140953) = 2 : 2.281906 3 ( 1 : 1.140953) = 3 : 3.422859 4 ( 1 : 1.140953) = 4 : 4.563812 5 ( 1 : 1.140953) = 5 : 5.704765 6 ( 1 : 1.140953) = 6 : 6.845718 7 ( 1 : 1.140953) = 7 : 7.986671 8 ( 1 : 1.140953) = 8 : 9.127624 That 7.986671 in row 7 looks extremely close to 8. I doubt I'd get much closer unless I go to extremely high integers. So it looks like the empirical formula for toluene is C7H8</span>

7 0

3 years ago

Excess magnesium reacts with 165.0 grams of hydrochloric acid in a single displacement reaction.

JulsSmile [24]

Answer:

The volume of hydrogen gas produced will be approximately 50.7 liters under STP.

Explanation:

Relative atomic mass data from a modern periodic table:

H: 1.008;
Cl: 35.45.

Magnesium is a reactive metal. It reacts with hydrochloric acid to produce

Hydrogen gas $\rm H_2$ , and
Magnesium chloride, which is a salt.

The chemical equation will be something like

$\rm ?\;Mg\;(s) + ?\;HCl \;(aq)\to ?\;H_2 \;(g)+ [\text{Formula of the Salt}]$ ,

where the coefficients and the formula of the salt are to be found.

To determine the number of moles of $\rm H_2$ that will be produced, first find the formula of the salt, magnesium chloride.

Magnesium is a group 2 metal. The oxidation state of magnesium in compounds tends to be +2.

On the other hand, the charge on each chloride ion is -1. Each magnesium ion needs to pair up with two chloride ions for the charge to balance in the salt, magnesium chloride. The formula for the salt will be $\rm MgCl_2$ .

$\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ ?\;MgCl_2\;(aq)$ .

Balance the equation. $\rm MgCl_2$ contains the largest number of atoms among all species in this reaction. Start by setting its coefficient to 1.

$\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ {\bf 1\;MgCl_2}\;(aq)$ .

The number of $\rm Mg$ and $\rm Cl$ atoms shall be the same on both sides. Therefore

$\rm {\bf 1\;Mg}\;(s) + {\bf 2\;HCl}\;(aq) \to ?\;H_2 \;(g)+ {1\;\underset{\wedge}{Mg}\underset{\wedge}{Cl_2}}\;(aq)$ .

The number of $\rm H$ atoms shall also conserve. Hence the equation:

$\rm {1\;Mg}\;(s) + {2\;\underset{\wedge}{H}Cl}\;(aq) \to {\bf 1\;H_2 \;(g)}+ {1\;MgCl_2}\;(aq)$ .

How many moles of HCl are available?

$M(\rm HCl) = 1.008 + 35.45 = 36.458\;g\cdot mol^{-1}$ .

$\displaystyle n({\rm HCl}) = \frac{m(\text{HCl})}{M(\text{HCl})} = \rm \frac{165.0\;g}{36.458\;g\cdot mol^{-1}} = 4.52576\;mol$ .

How many moles of Hydrogen gas will be produced?

Refer to the balanced chemical equation, the coefficient in front of $\rm HCl$ is 2 while the coefficient in front of $\rm H_2$ is 1. In other words, it will take two moles of $\rm HCl$ to produce one mole of $\rm H_2$ . $\rm 4.52576\;mol$ of $\rm HCl$ will produce only one half as much $\rm H_2$ .

Alternatively, consider the ratio between the coefficient in front of $\rm H_2$ and $\rm HCl$ is:

$\displaystyle \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}$ .

$\displaystyle n(\text{H}_2) = n(\text{HCl})\cdot \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}\;n(\text{HCl}) = \rm \frac{1}{2}\times 4.52576\;mol = 2.26288\;mol$ .

What will be the volume of that many hydrogen gas?

One mole of an ideal gas occupies a volume of 22.4 liters under STP (where the pressure is 1 atm.) On certain textbook where STP is defined as $\rm 1.00\times 10^{5}\;Pa$ , that volume will be 22.7 liters.

$V(\text{H}_2) = \rm 2.26288\;mol\times 22.4\;L\cdot mol^{-1} = 50.69\; L$ , or

$V(\text{H}_2) = \rm 2.26288\;mol\times 22.7\;L\cdot mol^{-1} = 51.37\; L$ .

The value "165.0 grams" from the question comes with four significant figures. Keep more significant figures than that in calculations. Round the final result to four significant figures.

5 0

2 years ago

Choose the false statement regarding resistance

Dmitry_Shevchenko [17]

I would say all if no statements are available

4 0

2 years ago

If the volume of a gas container at 32 degrees Celsius changes from 1.55 L to 755 mL, what will the final temperature be?

QveST [7]

So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K

4 0

2 years ago

The molecular weight of a gas is ________ g/mol if 3.5 g of the gas occupies 2.1 l at stp

bija089 [108]

<span>Pre-1982 definition of STP: 37 g/mol Post-1982 definition of STP: 38 g/mol This problem is somewhat ambiguous because the definition of STP changed in 1982. Prior to 1982, the definition was 273.15 K at a pressure of 1 atmosphere (101325 Pascals). Since 1982, the definition is 273.15 K at a pressure of exactly 100000 Pascals). Because of those 2 different definitions, the volume of 1 mole of gas is either 22.414 Liters (pre 1982 definition), or 22.71098 liters (post 1982 definition). And finally, there's entirely too many text books out there that still use the 35 year obsolete definition. So let's solve this problem using both definitions and you need to pick the correct answer for the text book you're using. First, determine how many moles of gas you have. Just simply divide the volume you have by the molar volume. Pre-1982: 2.1 / 22.414 = 0.093691443 moles Post-1982: 2.1 / 22.71098 = 0.092466287 moles Now determine the molar mass. Simply divide the mass by the moles. So Pre-1982: 3.5 g / 0.093691443 moles = 37.35666667 g/mol Post-1982: 3.5 g / 0.092466287 moles = 37.85163333 g/mol Finally, round to 2 significant figures. So Pre-1982: 37 g/mol Post-1982: 38 g/mol</span>

5 0

3 years ago