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2 years ago

Will give brainlist need answers asap

Chemistry

1 answer:

Serga [27]2 years ago

3 0

Answer:

40 g

Explanation:

Find the line labeled KClO3 (which might take you a min, theres a lot of lines here)

Notice that when the line creates a direct point, you can measure the exact temperature needed to dissolve a certain amount (like how they gave 30 degrees and it lined up perfectly with the 10 g line. )

Since its asking for the amount at 80 degrees, all you need to do is trace the line to the 80 degree point, and look at the grams. (notice it made a direct point, so there definitely should be any decimals or guesswork)

By reading the graph, you can tell that at 80 degrees, it dissolves 40 grams, and that is your answer.

Hope this helps :)

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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

Answer: The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

Explanation:

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )

Citrus2011 [14]

Answer:

a) ΔGrxn = 6.7 kJ/mol

b) K = 0.066

c) PO2 = 0.16 atm

Explanation:

a) The reaction is:

M₂O₃ = 2M + 3/2O₂

The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

$deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol$

b) To calculate the constant we have the following expression:

$lnK=-\frac{deltaG_{rxn} }{RT}$

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

$lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066$

c) The equilibrium pressure of O₂ over M is:

$K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm$

3 0

3 years ago

Water's heat of fusion is 80. cal/g , its specific heat is 1.0calg⋅∘C, and its heat of vaporization is 540 cal/g . A canister is

Pani-rosa [81]

294400 cal The heating of the water will have 3 phases 1. Melting of the ice, the temperature will remain constant at 0 degrees C 2. Heating of water to boiling, the temperature will rise 3. Boiling of water, temperature will remain constant at 100 degrees C So, let's see how many cal are needed for each phase. We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion. 80 cal/g * 320 g = 25600 cal Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C 420 * 100 = 42000 cal Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away. 420 g * 540 cal/g = 226800 cal So the total number of cal used is 25600 cal + 42000 cal + 226800 cal = 294400 cal

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3 years ago

Why are scientists worried that climate change will cause these toxic algae blooms to become more frequent?

ipn [44]

Answer: Eutrophication is the enhancement of the growth of algae in the water body.

Explanation:

The scientists are worried for the climate change as if the climate changed to prolonged rainy then the frequent raining can remove toxic chemicals from the agricultural sites, landfills, industries, and from other locations and deposit them to the water body (river, lakes, ponds, and others). The deposition of the salts of nitrogen, phosphorus, and sulfur promotes the growth of algae in the water body. This leads to reduction in the concentration of oxygen in the water body. This is called eutrophication. The lack of oxygen can lead to mortality of aquatic animals.

4 0

2 years ago

How many grams of sodium hydroxide are needed to make 500mLs of a 0.125M NaOH solution?

zubka84 [21]

Hope this helps you.

8 0

3 years ago