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IrinaVladis [17]

2 years ago

13

At what time of day would you be most likely to find that the air over water is significantly warmer than the air over land near

by?
A. 5:00 AM
B. 11:00 AM
C. 4:00 PM
D. 9:00 PM.

1 answer:

victus00 [196]2 years ago

5 0

A. 5:00 AM hope that helps!

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What is a good indicator for titrating potassium hydroxide with hydrobromic acid?

Cerrena [4.2K]

Potassium hydroxide is a strong base and hydrobromic acid is a strong acid. This implies that the pH of the end-point [neutralization] of their titration will be around pH 7. A good indicator for this kind of pH is bromthymol blue. This is because this indicator changes its colour at pH 7.

4 0

3 years ago

Manganese(iv) oxide reacts with aluminum to form elemental manganese and aluminum oxide: 3mno2+4al→3mn+2al2o3part awhat mass of

Oxana [17]

<span>12.4 g First, calculate the molar masses by looking up the atomic weights of all involved elements. Atomic weight manganese = 54.938044 Atomic weight oxygen = 15.999 Atomic weight aluminium = 26.981539 Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol Now determine the number of moles of MnO2 we have 30.0 g / 86.936044 g/mol = 0.345081265 mol Looking at the balanced equation 3MnO2+4Alâ†’3Mn+2Al2O3 it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So 0.345081265 mol / 3 * 4 = 0.460108353 mol So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum. 0.460108353 mol * 26.981539 g/mol = 12.41443146 g Finally, round to 3 significant figures, giving 12.4 g</span>

7 0

3 years ago

In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to

labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of $NH_3$ = 100 g

Molar mass of $NH_3$ = 27 g/mole

Molar mass of $N_2$ = 28 g/mole

First we have to calculate moles of $NH_3$ .

$\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles$

The given balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the given reaction, we conclude that

2 moles of $NH_3$ produced from 1 mole of $N_2$

3.7 moles of $NH_3$ produced from $\frac{1mole}{2mole}\times 3.7mole=1.85moles$ of $N_2$

Now we have to calculate the mass of $N_2$ .

Mass of $N_2$ = Moles of $N_2$ × Molar mass of $N_2$

Mass of $N_2$ = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

5 0

3 years ago

What is the molality of a solution made by dissolving 137.9g of sucrose in 414.1g of water?

Daniel [21]

Answer: 2.71 moles of solute for every 1 kg of solvent.

Explanation: As you know, the molality of a solution tells you the number of moles of solute present for every 1 kg of the solvent.This means that the first thing that you need to do here is to figure out how many grams of water are present in your sample. To do that, use the density of water.500.mL⋅1.00 g1mL=500. g Next, use the molar mass of the solute to determine how many moles are present in the sample.115g⋅1 mole NanO385.0g=1.353 moles NaNO3So, you know that this solution will contain 1.353moles of sodium nitrate, the solute, for 500. g of water, the solvent.In order to find the molality of the solution, you must figure out how many moles of solute would be present for 1 kg=103g of water.103g water⋅1.353 moles NaNO3500.g water=2.706 moles NaNO3You can thus say that the molality of the solution is equal to molality=2.706 mol kg−1≈2.71 mol kg−1 The answer is rounded to three sig figs.

8 0

3 years ago

A solution with [ OH] of 5 x 10-3 has a pH of

CaHeK987 [17]

Answer:

11·699

Explanation:

Given the concentration of hydroxide ion in the solution is 5 × $10^{-3}$ M

Assuming the temperature at which it is asked to find the pH of the solution be 298 K

<h3>At 298 K the dissociation constant of water is $10^{-14}$ </h3><h3>∴ pH + pOH = 14 at 298 K</h3><h3>pOH of the solution = -log( concentration of hydroxide ion )</h3>

∴ pOH of the given solution = - log(5 × $10^{-3}$ = -0·699 + 3 = 2·301

pH of the given solution = 14 - 2·301 = 11·699

∴ pH of the solution = 11·699

8 0

3 years ago