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lisov135 [29]
3 years ago
15

Use the mass and volume data to calculate the density of an unknown metal to the nearest hundredth mass of unknown metal = 222.5

0g volume of unknown metal = 25.00cm3 what is the density of the unknown metal

Chemistry
2 answers:
nasty-shy [4]3 years ago
4 0

Answer: D=8900g/L

Explanation:

Density is mass/volume. Mass in in grams and volume is in liters.

1 cm³=1 mL

With this conversion, we can convert cm³ to L.

D=\frac{222.50g}{25.00cm^3} *\frac{1cm^3}{1mL} *\frac{1000mL}{1L} =\frac{8900g}{L}

When we convert, we cancel out cm³ and mL to get L alone. The density is 8900 g/L.

aniked [119]3 years ago
4 0
All done no worries

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Answer:

John Dalton

Explanation:

Both John Dalton and Democritus thought that the atom was an indivisible sphere until J.J. Thompson came out with the plum pudding model. Hope I helped!

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3 years ago
What is the simplest formula of a compound if a sample of the compound contains 0.221 mol X, 0.442 mol Y, and 0.884 mol Z? chemP
Gnoma [55]

Answer:

XY₂Z₄

2.35 mol Z

Explanation:

A sample of the compound contains 0.221 mol X, 0.442 mol Y, and 0.884 mol Z. We can find the simplest formula (empirical formula) by <em>dividing all the numbers of moles by the smallest one</em>.

X: 0.221/0.221 = 1

Y: 0.442/0.221 = 2

Z: 0.884/0.221 = 4

The simplest formula is XY₂Z₄.

The molar ratio of X to Z is 1:4. The moles of Z in a sample that contained 0.588 moles of X is:

0.588 mol X × (4 mol Z/1 mol X) = 2.35 mol Z

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The levels, from smallest to largest, are: molecule, cell, tissue, organ, organ system, organism, population, community, ecosystem, biosphere.

Explanation:

The biological levels of organization of living things arranged from the simplest to most complex are: organelle, cells, tissues, organs, organ systems, organisms, populations, communities, ecosystem, and biosphere.

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What generic products are produced by an Acid/Base reaction?
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At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
erastova [34]

Answer:

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

3 0
3 years ago
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