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3 years ago

1. If I have 45 L of He in a balloon at 25 degrees celsius and increase the temperature of the

1 answer:

7 0

Use Charles' Law: V1/T1 = V2/T2. We assume the pressure and mass of the helium is constant. The units for temperature must be in Kelvin to use this equation (x °C = x + 273.15 K).

We want to solve for the new volume after the temperature is increased from 25 °C (298.15 K) to 55 °C (328.15 K). Since the volume and temperature of a gas at a constant pressure are directly proportional to each other, we should expect the new volume of the balloon to be greater than the initial 45 L.

Rearranging Charles' Law to solve for V2, we get V2 = V1T2/T1.

(45 L)(328.15 K)/(298.15 K) = 49.5 ≈ 50 L (if we're considering sig figs).

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Why reactivity of elements in group 1 increase from lithium to francium?

labwork [276]

Answer:

As you go down group 1, the number of electron shells increases – lithium has two, sodium has three, and so forth. The attraction from the positive nucleus to the negative electron is less. This makes it easier to remove the electron and makes the atom more reactive.

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3 years ago

Read 2 more answers

The heat of a reaction may be found with the equation q=mc deltaT.

ANEK [815]

Answer:Q=mcΔT Q = mc Δ T , where Q is the symbol for heat transfer, m is the mass of the substance, and ΔT is the change in temperature. The symbol c stands for specific heat and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00ºC.

Explanation:

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3 years ago

Calculate the [H+] and pH of a 0.0040 M hydrazoic acid solution. Keep in mind that the Ka of hydrazoic acid is 2.20×10−5. Use th

Vera_Pavlovna [14]

Answer:

$[H^+]=0.000285$

$pH=3.55$

Explanation:

In this, we can with the ionization equation for the hydrazoic acid ( $HN_3$ ). So:

$HN_3~~H^+~+~N_3^-$

Now, due to the Ka constant value, we have to use the whole equilibrium because this is not a strong acid. So, we have to write the Ka expression:

$Ka=\frac{[H^+][N_3^-]}{[HN_3]}$

For each mol of $H^+$ produced we will have 1 mol of $N_3^-$ . So, we can use "X" for the unknown values and replace in the Ka equation:

$Ka=\frac{X*X}{[HN_3]}$

Additionally, we have to keep in mind that $HN_3$ is a reagent, this means that we will be consumed. We dont know how much acid would be consumed but we can express a subtraction from the initial value, so:

$Ka=\frac{X*X}{0.004-X}$

Finally, we can put the ka value and solve for "X":

$2.2X10^-^5=\frac{X*X}{0.004-X}$

$2.2X10^-^5=\frac{X^2}{0.004-X}$

$X= 0.000285$

So, we have a concentration of 0.000285 for $H^+$ . With this in mind, we can calculate the pH value:

$pH=-Log[H^+]=-Log[0.000285]=3.55$

I hope it helps!

8 0

3 years ago

Please help....

stiv31 [10]

Hey there!:

Molarity HCl = 3.5 M

Volume HCl = ?

Molarity NaOH = 2.0 M

Volume NaOH in liters = 50.0 mL / 1000 => 0.05 L

Number of moles NaOH:

n = M * V

n = 2.0 * 0.05

n = 0.1 moles of NaOH

Given the reaction:

HCl + NaOH = NaCl + H2O

1 mole HCl --------- 1 mole NaOH

1 mol HCl reacts with 1 mol NaOH , so moles NaOH = moles HCl

0.1 moles of NaOH = 0.1 moles of HCl

Therefore:

M( HCl ) = n / V

3.5 = 0.1 / V

V = 0.1 / 3.5

V = 0.029 L in mL : 0.029 * 1000 => 29.0 mL

Answer B

hope that helps!

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3 years ago

What is parthenolide

vovangra [49]

Parthenolide is a sesquiterpene lactone which occurs naturally in the plant feverfew--highly concentrated in the flowers and fruit.

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3 years ago