Help me as soon as possible I’m gonna dieeee

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Answer 1

Answer: I think it’s Boiling point but I’m not sure

Answer 2

Answer: I think it’s D I’m not sure .........

Related Questions

A solid is hard brittle and electrically nonconducting. it's melt ( the liquid form of the substance) and an aqueous solution containing the substance conduct electricity. classify solid.
Which factor is likely to impact the possible number of compounds?
Please ignore my answers because they are wrong.
What you understand by macrocylic effect in coordination chemistry
The enthalpy of combustion of acetylene C2H2 is described by C2H2(g) + (5/2) O2(g) →2CO2(g) + H2O(l), Hrxn= –1299 kJ/mol. Calculate the enthalpy of formation of acetylene, given the following enthalpies of formation Hf[CO2(g)] = –393.5 kJ/mol Hf[H2O(l)] = –285.8 kJ/mo

The solubility of glucose at 30°C is125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water at 30°C. Explain your classification, and describe how you could increase the amount of glucose in the solution without adding more glucose.

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Answer:

Saturated solution

We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.

Explanation:

Step 1: Calculate the mass of water

The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.

$400 mL * (0.996g)/(1mL) =398g$

Step 2: Calculate the mass of glucose per 100 g of water

550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.

$100gH_2O * (550gGlucose)/(398gH_2O) = 138 gGlucose$

Step 3: Classify the solution

The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.

The solution made by adding 550 g of glucose to 400 mL of water at 30°C is saturated. If you want to increase the amount of glucose in the solution without adding more glucose, you can increase the temperature.

The solution made by adding 550 g of glucose to 400 mL of water at 30°C is saturated.

Since the solubility of glucose at 30°C is 125 g/100 g water, adding 550 g of glucose to 400 mL of water exceeds the maximum amount of glucose that can dissolve in the given amount of water.

To increase the amount of glucose in the solution without adding more glucose, you can increase the temperature. Higher temperatures generally increase the solubility of solutes in water. By increasing the temperature, you can dissolve more glucose in the solution.

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Given the reaction below, how will the concentration of D change if the concentration of B decreases by 0.045 M? 3A(g)+2B(g)⇋2C(g)+5D(g)

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Answer:

0.113 M

Explanation:

Since B and D are on opposite sides of the reaction, the concentration of D increases when the concentration of B decreases. The amount by which D increases is determined by the coefficients of B and D in the balanced chemical equation:

[D]= $(5)/(2)$ (0.045 M)=0.113 M.

Order the materials from causing the greatest refraction to causing the least refraction.A
water, air, oil

B
air, water, oil

C
oil, water, air

D
none of the above

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I think A. If not I’m sorryyy

Hi I wanna is a little bit more

How many water molecules are in a block of ice containing 1.50 mol of water (H2O)?

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1.50 moles H20(6.02*10^23 molecules/1 mole)
9.03*10^23 molecules

Final answer:

The number of water molecules in a 1.50 mol block of ice is calculated by multiplying the number of moles of water by Avogadro's number. The result is approximately 9.033 x 10^23 water molecules.

Explanation:

In chemistry, the amount of substance in moles is related to the number of particles (atoms, molecules) through Avogadro's number. Avogadro's number, which is 6.022 x 1023 particles/mol, tells us the number of molecules in one mole of a substance.

To calculate the number of water molecules in 1.50 mol of water, you would multiply the number of moles of water by Avogadro's number:

1.50 mol of water x 6.022 x 1023 water molecules/mol of water = 9.033 x 1023 water molecules

Therefore, there are approximately 9.033 x 1023 water molecules in a 1.50 mol block of ice.

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In a climatological sense, dryness is a function of both annual rainfall and ________.This a geology question.

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Answer:

In a climatological sense, dryness is a function of both annual rainfall and evaporation

Find the equilibrium constants, Kp, for the following equilibria, (i) NO(g) + ½ O2(g) ⇄ NO2(g), Kp = ? (ii) NO2(g) ⇄ NO(g) + ½ O2(g), Kp = ?, given the equilibrium constant, Kp, for the reaction: 2NO (g) + O2(g) ⇄ 2NO2(g) Kp= 100 at the same temperature

Answers

Answer :

(i) The value of equilibrium constants for this reaction is, 10

(ii) The value of equilibrium constants for this reaction is, 0.1

Explanation :

The given equilibrium reaction is,

$2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)$ $K_p=100$

Now we have to determine the equilibrium constants for the following equilibrium reactions.

(i) $NO(g)+(1)/(2)O_2(g)\rightleftharpoons NO_2(g)$ $K_p_1=?$

From the given reaction we conclude that, the reaction (i) will takes place when the given main reaction will be multiplied by half (1/2). That means when reaction will be half then the equilibrium constant will be:

$K_p_1=(K_p)^{(1)/(2)}$

$K_p_1=(100)^{(1)/(2)}$

$K_p_1=10$

The value of equilibrium constants for this reaction is, 10

(ii) $NO_2(g)\rightleftharpoons NO(g)+(1)/(2)O_2(g)$ $K_p_2=?$

From the given reaction we conclude that, the reaction (ii) will takes place when the reaction (i) will be reverse. That means when reaction will be reverse then the equilibrium constant will be:

$K_p_2=(1)/((K_p_1))$

$K_p_2=(1)/((10))$

$K_p_2=0.1$

The value of equilibrium constants for this reaction is, 0.1