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elena-s [515]
3 years ago
10

How many degrees is the earth tilted on its axis

Chemistry
2 answers:
otez555 [7]3 years ago
6 0

Answer: 2.35

Because sometimes the earth tilts change during cycle

kiruha [24]3 years ago
4 0

Answer:

23.5 degrees

Explanation:

Today, the Earth's axis is tilted 23.5 degrees from the plane of its orbit around the sun. But this tilt changes. During a cycle that averages about 40,000 years, the tilt of the axis varies between 22.1 and 24.5 degrees. Because this tilt changes, the seasons as we know them can become exaggerated.

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How many atoms are there in 2.3 grams of zirconiumn?
Effectus [21]

Answer:

1.51834×10^22

Explanation:

Number of atom=Mass/Molar Mass × avogadros constant

2.3g/91.224g/mol × 6.203×10^23

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3 years ago
Which organelle prepares material for storage within the cell
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The golgi apparatus prepares material for storage and processes them.
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Why is sulphur more reactive than chlorine​
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Because the most reactive elements are<em> HALOGENS</em>.

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What is the solution to the problem expressed to the correct number of significant figures
quester [9]

I don't know what the problem is, but here are some rues to help you out:

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  2. When a zero falls between non-zero digits, that zero is significant.
  3. When a zero falls after a decimal point, that zero is significant.
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4 0
4 years ago
What is the OH- of {H+} = 4.0 x 10 to the power of -8
True [87]

Answer:

At standard room temperature, [{\rm OH^{-}] \approx 2.5 \times 10^{-7}\; \rm M when [{\rm H^{+}] = 4.0 \times 10^{-8}\; \rm M.

Explanation:

The following equilibrium goes on in water:

{\rm H_{2}O}\, (l) \rightleftharpoons {\rm H^{+}}\, (aq) + {\rm OH^{-}}\, (aq).

The forward reaction is known as the self-ionization of water. The ionization constant of water, K_{\rm w}, gives the equilibrium position of this reaction:

K_{\rm w} = [{\rm H^{+}] \cdot [{\rm OH^{-}}].

At standard room temperature (25\; {\rm ^{\circ}C}), K_{\rm w} \approx 10^{-14}. Also, [{\rm H^{+}}] = 4.0 \times 10^{-8}\; \rm mol \cdot L^{-1}. Substitute both values into the equation and solve for [{\rm OH^{-}}].

\begin{aligned} {[}{\rm OH^{-}}{]} &= \frac{K_{\rm w}}{[{\rm H^{+}}]} \\ &\approx \frac{10^{-14}}{4.0 \times 10^{-8}} = 2.5 \times 10^{-7}\end{aligned}.

In other words, in an aqueous solution at standard room temperature, [{\rm OH^{-}] \approx 2.5 \times 10^{-7}\; \rm M when [{\rm H^{+}] = 4.0 \times 10^{-8}\; \rm M.

5 0
3 years ago
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