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anygoal [31]
3 years ago
5

At which type of tectonic plate boundary is a volcano least likely to occur?

Chemistry
2 answers:
andrey2020 [161]3 years ago
8 0

Answer: the answer is a

Explanation:

I took the test and got it wrong because someone said the answer was divergent boundary and it’s wrong. I hate them I hate them I hate them. I hate it I hate school

Katena32 [7]3 years ago
7 0

Answer:

A divergent boundary is when the plates move apart from each other. When the plates part, magma from under either plate rises and forms a volcano. A hotspot is the third place a volcano can form. This particular type is the least common.

Explanation:

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How fast do i have to run 250 m in 30 seconds
elena-14-01-66 [18.8K]

Answer:

S=D/T

250/30= 8.33 miles

Explanation:

7 0
3 years ago
How many mL of 0.013 M potassium hydroxide are required to reach the equivalence point in the titration of 75 mL 0.166 M hydrocy
umka21 [38]

Answer:

957.7mL

Explanation:

Using the formula below;

CaVa = CbVb

Where;

Ca = concentration of acid (M)

Va = volume of acid (mL)

Cb = concentration of base (M)

Vb = volume of base (mL)

According to the information provided in this question:

Ca = 0.166 M

Cb = 0.013 M

Va = 75mL

Vb = ?

Using CaVa = CbVb

0.166 × 75 = 0.013 × Vb

12.45 = 0.013Vb

Vb =12.45/0.013

Vb = 957.7mL

6 0
3 years ago
What is oxidation state?​
solniwko [45]

Answer:

a number assigned to an element in chemical combination which represents the number of electrons lost (or gained, if the number is negative), by an atom of that element in the compound

Explanation:

sodium, magnesium, iron is zero.

4 0
3 years ago
Read 2 more answers
Match these items with their examples.
kiruha [24]
Here you go! Feel free to ask questions!

7 0
4 years ago
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A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

6 0
3 years ago
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