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Ray Of Light [21]
2 years ago
10

Use the drop-down menus to label each of the following changes P for physical change and C for the chemical change. The substanc

e changes to a new substance, the original substance can be recovered, the color changes, gas is produced and given off, and the substance changes size, shape, or volume
Chemistry
2 answers:
zhannawk [14.2K]2 years ago
8 0

Answer:c,p,c,c,p

Explanation:

dlinn [17]2 years ago
6 0

Answer:

Chemical, Physical, Chemical, Chemical, Physical!!

Explanation:

I just did it correctly.

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Please give me the answer please
zepelin [54]

Answer:

A. 30cm³

Explanation:

Based on the chemical reaction:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

<em>1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂</em>

<em />

To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:

<em>Moles CaCO₃ -Molar mass: 100.09g/mol-</em>

1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles

<em>Moles HCl:</em>

50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles

For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:

2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.

The moles produced of CO₂ are:

2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂

Using PV = nRT

<em>Where P is pressure = 1atm assuming STP</em>

<em>V volume in L</em>

<em>n moles = 1.25x10⁻³ moles CO₂</em>

<em>R gas constant = 0.082atmL/molK</em>

<em>T = 273.15K at STP</em>

<em />

V = nRT / P

1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V

0.028L = V

28cm³ = V

As 28cm³ ≈ 30cm³

Right option is:

<h3>A. 30cm³</h3>

5 0
3 years ago
What is the composition of Gilsonite?
nekit [7.7K]

Answer:

Roughly C100 H140 N3 O

Explanation:

Gilsonite is a bituminous product that resembles shiny black obsidian.

It contains more than 100 elements.

Its mass composition varies but is approximately 84 % C, 10 % H, 3 % N, and 1 % O.

Its empirical formula is roughly C100 H140 N3 O.

3 0
3 years ago
The term used to describe the rapid release of bubbles from a liquid is _____
aksik [14]
The term used to describe the rapid release of bubbles, or rapid release of a gas from a liquid or a solution is called Effervescence.  The bubbling of a solution is due to the escape of a gas which may be from a chemical reaction, as in fermenting liquid, or by coming out of a solution after having been under pressure, as in a carbonated drink. For example; soda, champagne among others.
5 0
3 years ago
An ionic bond can be formed when one or more electrons are:
Mashutka [201]
D - for example, Potassium has 1 electron on its outer shell, whilst Chlorine has 7 electrons on its outer shell. Potassium loses one electron to Chlorine so that each of them have a full outer shell. This would form Potassium Chloride.
6 0
2 years ago
Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, th
Gre4nikov [31]

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M

The given equilibrium reaction is,

                           N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

Initially conc.      0        0           0.1576

At eqm.               (x)       (x)        (0.1576-2x)

The expression of K_c will be,

K_c=\frac{[NO]^2}{[N_2][O_2]}

0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}

By solving the term, we get:

x=0.0742,0.0839

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.

5 0
3 years ago
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