Which of the following is true about limiting and excess reagents? Links are reported

Elements that want to gain one outer electron?

mario62 [17]

Answer:

halogens

Explanation:

Elements that are nonmetals tend to gain electrons and become negatively charged ions called anions

8 0

3 years ago

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Tumu’s class was given an assignment to feature a scientist that contributed to the development of the cell theory. The class de

xxMikexx [17]

Answer:

The types of cells at work in the human body

Explanation:

Rudolf Virchow was a German physician who is known as the father of modern pathology. He contributed to the development of the cell theory by concluding that all cells come from preexisting cells. This led to the conclusion that diseases occur at the cellular level, which allowed scientists to diagnose diseases more accurately.

Considering that Virchow analyzed human cells, the image that Tumu would be most likely to use in his assignment is the one of the types of cells at work in the human body.

4 0

3 years ago

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What volume does 1.5 x 10^24 molecules of CO2 occupy at STP?

Nookie1986 [14]

Answer:

The volume occupied by given molecules of carbon dioxide at STP is 55.86 L.

Explanation:

$N=n\times N_A$

Where:

N = Number of particles / atoms/ molecules

n = Number of moles

$N_A=6.022\times 10^{23} mol^{-1}$ = Avogadro's number

We have:

N = $1.5\times 10^{24}$

n =?

$n=\frac{N}{N_A}=\frac{1.5\times 10^{24}}{6.022\times 10^{23} mol^{-1}}$

n = 2.4909 moles

Using ideal gas equation:

PV = nRT

where,

P = Pressure of carbon dioxide gas = $1 atm$ (at STP)

V = Volume of carbon dioxide gas = ?

n = number of moles of carbon dioxide gas =2.4909 mol

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of carbon dioxide gas = 273.15 K (at STP)

Putting values in above equation, we get:

$V=\frac{2.4909 mol\times 0.0821 atm L/mol K\times 273.15 K}{1 atm}$

V = 55.86 L

The volume occupied by given molecules of carbon dioxide at STP is 55.86 L.

7 0

3 years ago

The acidic ingredient in vinegar is acetic acid. The pH of vinegar is around 2.4, and the molar concentration of acetic acid in

IrinaK [193]

Answer: The value of acid ionization constant $K_a$ for acetic acid is $1.87\times 10^{-5}$

Explanation:

$CH_3COOH\rightarrow H^+CH_3COO^-$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.85 M and $pH$ = 2.4

$pH=-log[H^+]$

$[H^=}=c\times \alpha=10^{-2.4}=3.98\times 10^{-3}$

$K_a=?$

Putting in the values we get:

$K_a=\frac{(3.98\times 10^{-3})^2}{(0.85-3.98\times 10^{-3})}=1.87\times 10^{-5}$

Thus the value of acid ionization constant $K_a$ for acetic acid is $1.87\times 10^{-5}$

5 0

3 years ago

A hydrate is compound that included water molecules within its crystal structure. During an experiment to determine the percent

stellarik [79]

Answer:

The percent by mass of water in this crystal is:

21.4%

Explanation:

This exercise can be easily solved using a simple rule of three where the initial weight of the hydrated crystal (6,235 g) is taken into account as 100% of the mass, and the percentage to which the mass of 4.90 g corresponds (after getting warm). First, the values and unknown variable are established:

6,235 g = 100%
4.90 g = X

And the value of the variable X is found:

X = (4.90 g * 100%) / 6,235 g
X = approximately 78.6%.

The calculated value is not yet the percentage of the water, since the water after heating the glass has evaporated, therefore, the remaining percentage must be taken, which can be calculated by subtraction:

Water percentage = Total percentage - Percentage after heating.
Water percentage = 100% - 78.6% = 21.4%

4 0

4 years ago

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