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3 years ago

The rising magma that may result from this type of plate movement may produce HELP

Chemistry

1 answer:

nirvana33 [79]3 years ago

7 0

Volcanic islands, seamounts, guyouts

Explanation:

The magma rising from this type of plate movement may produce under water volcanic activities.

The magma here would be mafic - ultramafic because it is produced from melting of oceanic plates.

As the subducting oceanic plate melts and begins to rise, it will settle and cool on the ocean floor resulting in the formation of massive and extensive seamounts.

With time, they outgrow the ocean level and forms volcanic Islands.

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lithosphere brainly.com/question/9582362

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2 years ago

A compound is found to contain 37.32 % phosphorus , 16.88 % nitrogen , and 45.79 % fluorine by

Alla [95]

Answer: 1. The empirical formula is $PNF_2$

2. The molecular formula is $PNF_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of P = 37.32 g

Mass of N = 16.88 g

Mass of F = 45.79 g

Step 1 : convert given masses into moles.

Moles of P = $\frac{\text{ given mass of P}}{\text{ molar mass of P}}= \frac{37.32g}{31g/mole}=1.20moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.88g}{14g/mole}=1.20moles$

Moles of F = $\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{45.79g}{19g/mole}=2.41moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For P = $\frac{1.20}{1.20}=1$

For N = $\frac{1.20}{1.20}=1$

For F = $\frac{2.41}{1.20}=2$

The ratio of P: N: F= 1: 1: 2

Hence the empirical formula is $PNF_2$

The empirical weight of $PNF_2$ = 1(31)+1(14)+2(19)= 82.98 g.

The molecular weight = 82.98 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{82.98}{82.98}=1$

The molecular formula will be= $1\times PNF_2=PNF_2$

3 0

3 years ago

A geochemist in the field takes a 36.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X.

NeTakaya

Answer:

solubility of X in water at 17.0 $^{0}\textrm{C}$ is 0.11 g/mL.

Explanation:

Yes, the solubility of X in water at 17.0 $^{0}\textrm{C}$ can be calculated using the information given.

Let's assume solubility of X in water at 17.0 $^{0}\textrm{C}$ is y g/mL

The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.

So, solubility of X in 1 mL of water = y g

Hence, solubility of X in 36.0 mL of water = 36y g

So, 36y = 3.96

or, y = $\frac{3.96}{36}$ = 0.11

Hence solubility of X in water at 17.0 $^{0}\textrm{C}$ is 0.11 g/mL.

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3 years ago