Why would it have been a bad idea to plant 100 seeds in your plant experiment?

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 28.0 g of aluminum?

Flauer [41]

Answer: 2Al(s) + 3Cl2(g) ---> 2AlCl3(s) Always work in moles moles = mass of substane / molar mass 2Al (s) + 3Cl (g) --> 2AlCl3 (s) moles Al = 23.0 g / 26.98 g/mol = 0.852 moles Al

5 0

3 years ago

Which of the following wave types has a wavelength closest to visible light?

Dominik [7]

I belive its b
hope this helps :)

3 0

3 years ago

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Suppose now that you wanted to determine the density of a small crystal to confirm that it is sulfur. From the literature, you k

ivolga24 [154]

Answer:

Mix 11.73 mL of CHCl₃ and 8.27 mL of CHBr₃

Explanation:

To begin, it seems your question lacks the name of the pure samples that will be mixed to prepare the liquid mixture. The names are not necessary, as the numerical values are there. However, an internet search tells me they are CHCl₃ (d=1.492g/mL) and CHBr₃ (d=2.890 g/mL).

The mixture has a density of 2.07 g/mL, so 20 mL of the mixture would weigh:

20 mL * 2.07 g/mL = 41.4 g

Let X be the volume of CHCl₃ and Y the volume of CHBr₃:

X + Y = 20 mL

The mass of CHCl₃ and CHBr₃ combined have to be equal to the mass of the mixture. We can write that equation using the volume of the samples and their density:

X * 1.492 + Y * 2.890 = 41.4 g

So now we have a system of two equations and two unknowns, we use algebra to solve it:

1. Express Y in terms of X:

X + Y = 20

Y = 20 - X

2. Replace Y in the second equation:

X * 1.492 + Y * 2.890 = 41.4

1.492*X + 2.890*(20-X) = 41.4

3. Solve for X:

1.492*X + 57.8 - 2.890*X = 41.4

1.398*X = 16.4

X = 11.73 mL

4. Using the now known value of X, solve for Y:

X + Y = 20

11.73 + Y = 20

Y = 8.27 mL

So, to prepare the liquid mixture we would mix 8.27 mL of CHBr₃ and 11.73 mL of CHCl₃.

7 0

4 years ago

If an aqueous solution of urea n2h4co is 26% by mass and has density of 1.07 g/ml, calculate the molality of urea in the soln

Ksju [112]

Answer is: molality of urea is 5.84 m.

If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.

5 0

3 years ago

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Match the following associations. Exothermic or endothermic

Alekssandra [29.7K]

Answer:

1) ∆H is positive $\Rightarrow$ Endothermic