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Vlad [161]
2 years ago
7

Imagine that you and your friend are at the beach. You have your chairs and umbrella set up in the perfect spot near the water s

o you can jump in to cool off with ease. A few hours into your stay, the water line has managed to creep up closer and closer to your chairs and your things start to get wet! Your friend points up at the Moon and says it is to blame. What does this mean? Write a response in complete sentences. I NEED A ANSWER PLZZZZZ! And this is a science question.
Chemistry
1 answer:
sineoko [7]2 years ago
8 0

Answer:

High tides and low tides are caused by the moon. The moon's gravitational pull generates something called the tidal force. The tidal force causes Earth—and its water—to bulge out on the side closest to the moon and the side farthest from the moon. ... When you're not in one of the bulges, you experience a low tide.

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A complex, ML₆²⁺, is violet. The same metal forms a complex with another ligand, Q, that creates a weaker field. What color migh
Bingel [31]

A complex, ML₆²⁺, is violet. The same metal forms a complex with another ligand, Q, that creates a weaker field.  MQ₆²⁺, be expected to show green color.

<h3>What is spectrochemical series?</h3>

The ligands (attachment to a metal ion) are listed in the spectrochemical series according to the strength of their field. The series has been created by superimposing several sequences discovered through spectroscopic research because it is impossible to produce the full series by examining complexes with a single metal ion. The halides are referred to be weak-field ligands whereas the ligands cyanide and CO are strong-field ligands. Medium field effects are claimed to be produced by ligands like water and ammonia.

To know more about the spectrochemical series, visit:

brainly.com/question/27892620

#SPJ4

4 0
1 year ago
In another experiment, a 0.150 M BF4^-(aq) solution is prepared by dissolving NaBF4(s) in distilled water. The BF4^-(aq) ions in
Ilia_Sergeevich [38]

Answer:

A) Forward rate = 1.1934 × 10^(-4) M/min

B) I disagree with the claim

Explanation:

A) We are told that [HF] reaches a constant value of 0.0174 M at equilibrium.

The reversible reaction given to us is;

BF4-(aq) +H20(l) → BF3OH-(aq) + HF(aq)

From this, we can see that the stoichiometric ratio is 1:1:1:1

Thus, concentration of [BF4-] is now;

[BF4-] = 0.150 - 0.0174

[BF4-] = 0.1326 M

From the rate law, we are told the forward rate is kf [BF4-].

We are given Kf = 9.00 × 10^(-4) /min

Thus;

Forward rate = 9.00 × 10^(-4) /min × (0.1326M)

Forward rate = 1.1934 × 10^(-4) M/min

(B) The student claims that the initial rate of the reverse reaction is equal to zero can't be true because at equilibrium, rates for the forward and reverse reactions are usually equal.

Thus, I disagree with the claim.

3 0
2 years ago
How many grams of co2 is produced when oxygen reacts with carbon?
Serggg [28]
44g of CO2 can produce by the reaction of carbon with oxygen
5 0
3 years ago
How many protons, electrons, and neutrons does the following isotope contain?<br> 27 A13+
Lapatulllka [165]

Hey there!

27 tells us the sum of protons and neutrons is 27.

Al tells us we have 13 protons.

3+ tells us that there are 3 less electrons than protons.

13 + n = 27

neutrons = 14

13 - 3 = 10 electrons

27Al3+ has 13 protons, 14 neutrons, and 10 electrons.

Hope this helps!

7 0
3 years ago
How much energy is required to raise the temperature of 10.6 grams of gaseous neon from
Alona [7]

Answer:

Approximately 1.95 \times 10^{2}\; \rm J.

Explanation:

Look up the specific heat of gaseous neon:

c = 1.03 \; \rm J \cdot g^{-1} \cdot K^{-1}.

Calculate the required temperature change:

\Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K.

Let m denote the mass of a sample of specific heat C. Energy required to raise the temperature of this sample by \Delta T:

Q = c \cdot m \cdot \Delta T.

For the neon gas in this question:

  • c = 1.03\; \rm J \cdot g^{-1}\cdot K^{-1}.
  • m = 10.6\; \rm g.
  • \Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K.

Calculate the energy associated with this temperature change:

\begin{aligned}Q &= c \cdot m \cdot \Delta T \\ &= 1.03\; \rm J \cdot g^{-1}\cdot K^{-1} \times 10.6\; \rm g \times 17.9\; \rm K \\ &\approx 1.95 \times 10^{2}\; \rm J\end{aligned}.

3 0
3 years ago
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