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3 years ago

You will be reading from an essay and a play dealing with the lives of women in the late eighteenth and early

Chemistry

2 answers:

s344n2d4d5 [400]3 years ago

6 0

Answer:

all work

Explanation:

Radda [10]3 years ago

4 0

Answer:

lol, this a warm up question, and the answer is all of them

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Calcium Carbonate decomposes at 1200°C to form carbon dioxide and

Travka [436]

This is a Charles' Law problem: V1/T1 = V2/T2. As the temperature of a fixed mass of gas decreases at a constant pressure, the volume of the gas should also decrease proportionally.

To use Charles' Law, the temperature must be in Kelvin (x °C = x + 273.15 K). We want to solve Charles' Law for V2, which we can obtain by rearranging the equation into V2 = V1T2/T1. Given V1 = 25 L, T1 = 1200 °C (1473.15 K), and T2 = 25 °C (298.15 K):

V2 = (25 L)(298.15 K)/(1473.15 K) = 5.1 L.

5 0

4 years ago

Electrolysis can be used for all of the following, except A. separating metals from ores. B. plating objects with metal coatings

Assoli18 [71]

The answer is D. recharging batteries.

3 0

4 years ago

249 g of potassium iodide, KI, is mixed with 496.5 g of lead(II) nitrate, Pb(NO3)2.

sveticcg [70]

Answer: The number of moles of excess reagent left is 0.750 moles

Explanation:

The number of moles is calculated by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

For KI:

Given mass of $KI$ = 249 g

Molar mass of $KI$ = 166 g/mol

Putting values in equation 1, we get:

$\text{Moles of KI}=\frac{249g}{166g/mol}=1.5mol$

For $Pb(NO_3)_2$ :

Given mass of $Pb(NO_3)_2$ = 496.5 g

Molar mass of $Pb(NO_3)_2$ = 331 g/mol

Putting values in equation 1, we get:

$\text{Moles of }Pb(NO_3)_2=\frac{496.5g}{331g/mol}=1.5mol$

The given chemical equation follows:

$2KI+Pb(NO_3)_2\rightarrow 2KNO_3+PbI_2$

By stoichiometry of the reaction:

If 2 moles of KI reacts with 1 mole of lead(II) nitrate

So, 1.5 moles of KI will react with = $\frac{1}{2}\times 1.5=0.75mol$ of lead(II) nitrate

As the given amount of lead(II) nitrate is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, KI is considered a limiting reagent because it limits the formation of the product.

Moles of excess reactant ( $Pb(NO_3)_2$ ) left = [1.5 - 0.75] = 0.750 moles

Hence, the number of moles of excess reagent left is 0.750 moles

8 0

3 years ago

Describe how the kinetic and potential energies of pendulum are changing at position B. Needed ASAP

zaharov [31]

The kinetic energy of the pendulum is increasing and the potential energy is decreasing. Potential energy is being converted to kinetic energy.

I hope this helps!

7 0

4 years ago

Read 2 more answers

What is the empirical formula for Hg2(NO3)2

natima [27]

I think it's HgNO3 because the subscripts on Hg and No3 are divisible by 2 :)

5 0

4 years ago