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mojhsa [17]
3 years ago
6

In which environment would primary succession occur?

Chemistry
2 answers:
OLEGan [10]3 years ago
4 0

an area of exposed rock after a glacier melts away

taurus [48]3 years ago
3 0

Answer:

an area of exposed rock after a glacier melts away

Explanation:

i took the test

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What is created when an acid is mixed with a base?
Arturiano [62]

Answer:

If you mix equal amounts of a strong acid and a strong base, the two chemicals essentially cancel each other out and produce a salt and water. Mixing equal amounts of a strong acid with a strong base also produces a neutral pH (pH = 7) solution.

8 0
3 years ago
Ibuprofen, a well‑known, non‑steroidal anti‑inflammatory drug, has chirality.
Liula [17]

Answer:

A (True)

Explanation:

Because ibuprofen has a chiral carbon center (carbon bonded to 4 distinct groups of atoms).

This means that a mixture of ibuprofen can rotate plane-polarized light equally in both the clockwise and counterclockwise direction.

8 0
2 years ago
Read 2 more answers
If a gas occupies 79.5 mL at -1.4°C, what temperature, in Kelvin, would it
Anna35 [415]

Answer:

121 K

Explanation:

Step 1: Given data

  • Initial volume (V₁): 79.5 mL
  • Initial temperature (T₁): -1.4°C
  • Final volume (V₂): 35.3 mL

Step 2: Convert "-1.4°C" to Kelvin

We will use the following expression.

K = °C + 273.15 = -1.4°C + 273.15 = 271.8 K

Step 3: Calculate the final temperature of the gas (T₂)

Assuming ideal behavior and constant pressure, we can calculate the final temperature of the gas using Charles' law.

V₁/T₁ = V₂/T₂

T₂ = V₂ × T₁/V₁

T₂ = 35.3 mL × 271.8 K/79.5 mL = 121 K

5 0
2 years ago
Identify the correct coefficients to balance the redox reaction with the lowest possible integer coefficients.
Monica [59]

Answer:

\rm 3\; Ag^{1+} + 1\; Al \to 1\; Al^{3+} + 3\; Ag.

Explanation:

Electrons are conserved in a chemical equation.

The superscript of \rm Ag^{1+} indicates that each of these ions carries a charge of +1. That corresponds to the shortage of one electron for each \rm Ag^{+} ion.

Similarly, the superscript +3 on each \rm Al^{3+} ion indicates a shortage of three electrons per such ion.

Assume that the coefficient of \rm Ag^{+} (among the reactants) is x, and that the coefficient of \rm Al^{3+} (among the reactants) is y.

\rm \mathnormal{x}\; Ag^{1+} + ?\; Al \to \mathnormal{y}\; Al^{3+} + ?\; Ag.

There would thus be x silver (\rm Ag) atoms and y aluminum (\rm Al) atoms on either side of the equation. Hence, the coefficient for \rm Al\! and \rm Ag\! would be y\! and x\!, respectively.

\rm \mathnormal{x}\; Ag^{1+} + \mathnormal{y}\; Al \to \mathnormal{y}\; Al^{3+} + \mathnormal{x}\; Ag.

The x \rm Ag^{1+} ions on the left-hand side of the equation would correspond to the shortage of x electrons. On the other hand, the y Al^{3+} ions on the right-hand side of this equation would correspond to the shortage of 3\, y electrons.

Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of x electrons, the right-hand side should also be x\! electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of 3\, y electrons. These two expressions should have the same value. Therefore, x = 3\, y.

The smallest integer x and y that could satisfy this relation are x = 3 and y = 1. The equation becomes:

\rm 3\; Ag^{1+} + 1\; Al \to 1\; Al^{3+} + 3\; Ag.

6 0
3 years ago
When the amount of oxygen is limited, carbon and oxygen react to form carbon monoxide. How many grams of CO can be formed from 3
Alinara [238K]

<u>61.25 grams</u> of CO can be formed from 35 grams of oxygen.

The molecular mass of oxygen is <u>16 gmol⁻¹</u>

The molecular mass of carbon monoxide is<u> 28 gmol⁻¹</u>

Explanation:

The molar mass of carbon monoxide is molar mass of C added to that of O;

12 + 16 = 28

= 28g/mol

The molar mass of oxygen is 16 g/mol while that of oxygen gas (O₂) is 32 g/mol

Since the ration oxygen to carbon monoxide is 1: 2 moles, we begin to find out how many moles of carbon monoxide are formed by 35 g of oxygen;

35/32 * 2

= 70/32 moles

Then multiply by the molar mass of carbon monoxide;

70/32 * 28

= 61.25 g

4 0
2 years ago
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