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melomori [17]
3 years ago
13

Can someone please help me, best answer gets BRAINLIEST

Chemistry
1 answer:
salantis [7]3 years ago
7 0

Answer:

9839 is your answer mate .

Plz mark brainliest and follow me ❤❤❤❤

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What would happen if the sand dunes in an area were destroyed?
Komok [63]

Answer:

<u>Our beaches would be unprotected</u>

In the short-term, these artificial sand hills will be destroyed by the elements. Because sand dunes protect inland areas from swells, tides, and winds, they must be protected and defended like national treasures. ... The ocean and the wind can have an unpredictable, destructive force on coastal regions.

- surfertoday

Natural sand dunes play a vital role in protecting our beaches, coastline and coastal developments from coastal hazards such as erosion, coastal flooding and storm damage. Sand dunes protect our shorelines from coastal erosion and provide shelter from the wind and sea spray.

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3 0
2 years ago
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Balance this equation. Not all spots have to have a number.
Lisa [10]

2Na+2H2O----2NaOH+H2

3 0
3 years ago
24. What volume of a 0.0200M calcium hydroxide is required to neutralize 35.00 mL of 0.0500M nitric acid
Natalka [10]

Answer:

THE VOLUME OF 0.200M CALCIUM HYDROXIDE NEEDED TO NEUTRALIZE 35 mL of 0.050 M NITRIC ACID IS 43.75 mL.

Explanation:

Using

Ca VA / Cb Vb = Na / Nb

Ca = 0.0500 M

Va = 35 mL

Cb = 0.0200 M

Vb = unknown

Na = 2

Nb = 1

Equation for the reaction:

Ca(OH)2 + 2HNO3 --------> Ca(NO3)2 + 2H2O

So therefore, we make Vb the subject of the equation and solve for it

Vb = Ca Va Nb / Cb Na

Vb = 0.0500 * 35 * 1 / 0.0200 * 2

Vb = 1.75 / 0.04

Vb = 43.75 mL

The volume of 0.02M calcium hydroxide required to neutralize 35 mL of 0.05 M nitric acid is 43.75 mL

6 0
3 years ago
Need help ASAP... (Question from 'Observing the Bunsen Burner'). "The match head is supposed to stay unlit. What does this tell
Solnce55 [7]

Answer:

It is too hot.

Explanation:

7 0
2 years ago
34. 3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases
Digiron [165]

Answer:

ΔH = 2.68kJ/mol

Explanation:

The ΔH of dissolution of a reaction is defined as the heat produced per mole of reaction. We have 3.15 moles of the solid, to find the heat produced we need to use the equation:

q = m*S*ΔT

<em>Where q is heat of reaction in J,</em>

<em>m is the mass of the solution in g,</em>

<em>S is specific heat of the solution = 4.184J/g°C</em>

<em>ΔT is change in temperature = 11.21°C</em>

The mass of the solution is obtained from the volume and the density as follows:

150.0mL * (1.20g/mL) = 180.0g

Replacing:

q = 180.0g*4.184J/g°C*11.21°C

q = 8442J

q = 8.44kJ when 3.15 moles of the solid react.

The ΔH of the reaction is:

8.44kJ/3.15 mol

= 2.68kJ/mol

5 0
3 years ago
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