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3 years ago

A nickel is placed in a cup of boiling water. Predict whether a solution will form.

Chemistry

1 answer:

Volgvan3 years ago

6 0

I think the copper could melt to cup and the water would smell like rust

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A chemical process dissolves 500 milligrams of iron oxide every 20 minutes. How long would it take this reaction to dissolve 2 l

stiv31 [10]

Answer:

36290 min = 604.8 hr.

Explanation:

Knowing that:

1 lbs = 453.59237 grams.

∴ 2 lbs = 907.18474 grams.

<em><u>Using cross multiplication:</u></em>

500 mg of iron oxide dissolved → 20 minutes.

907184.74 mg of iron oxide dissolved → ??? minutes.

<em>∴ The time needed to dissolve 2 lbs of iron oxide =</em> (907184.74 mg)(20 min)/(500 mg) = <em>36290 min = 604.8 hr.</em>

4 0

3 years ago

Which group of elements shares characteristics with both metals and nonmetals?

sertanlavr [38]

The answer to your question is,

Metalloids. They are a mix of elements that are both metals and non-metals in one.

-Mabel <3

4 0

3 years ago

Read 2 more answers

2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

No. of moles of CO = 0.800 mol

No. of moles of $H_2$ = 2.40 mol

Volume = 8.00 L

Concentration = $\frac{Moles}{Volume\ in\ L}$

Concentration of CO = $\frac{0.800}{8.00} = 0.100\ mol/L$

Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

$Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90$

8 0

3 years ago

What is the pressure of 1.50 moles of a gas in a 30.0L tank at a temperature of<br> 285K?

julia-pushkina [17]

The pressure of the gas : 1.1685 atm

<h3>Further explanation</h3>

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.08206 L.atm / mol K

T = temperature, Kelvin

n=moles=1.5

V=volumes = 30 L

T=temperature=285 K

The pressure :

$\tt P=\dfrac{nRT}{V}\\\\n=\dfrac{1.5\times 0.082\times 285}{30}\\\\P=1.1685~atm$

7 0

2 years ago

While working in the lab, you observe the following volume.

MakcuM [25]

Answer:

B 18.4L

Explanation:

7 0

2 years ago

Read 2 more answers