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V125BC [204]
4 years ago
13

During metamorphic processes, increased pressure and temperature can affect the _______ of minerals in rock. Rocks subjected to

very high pressure are typically _______ than others because mineral grains are squeezed together, and the atoms are more closely packed. During metamorphic processes, water facilitates the transfer of ions between and within minerals, which can _______ the rate at which metamorphic reactions take place. The growth of new minerals within a rock during metamorphism has been estimated to be about _______ per million years. _______ metamorphism is commonly associated with convergent plate boundaries, where two plates move toward each other. During contact metamorphism, a large intrusion will contain _______ thermal energy and will cool much more slowly than a small one. Metamorphosed sandstone is known as _______. The metamorphic rock _______, made from metamorphosed shale, was once used to make blackboards for classrooms.
Chemistry
2 answers:
Temka [501]4 years ago
6 0

Answer:

Stability , Denser,  Increase, 1 millimetre, Regional, More, Quartzite, Slate

Explanation:

During metamorphic processes, increased pressure and temperature can affect the <u>stability</u> of minerals in rock. Rocks subjected to very high pressure are typically <u>denser</u> than others because mineral grains are squeezed together, and the atoms are more closely packed. During metamorphic processes, water facilitates the transfer of ions between and within minerals, which can <u>increase</u> the rate at which metamorphic reactions take place. The growth of new minerals within a rock during metamorphism has been estimated to be about <u>1 millimeter</u> per million years. <u>Regional</u> metamorphism is commonly associated with convergent plate boundaries, where two plates move toward each other. During contact metamorphism, a large intrusion will contain <u>more</u> thermal energy and will cool much more slowly than a small one. Metamorphosed sandstone is known as <u>quartzite</u>. The metamorphic rock <u>slate</u>, made from metamorphosed shale, was once used to make blackboards for classrooms.

Stells [14]4 years ago
4 0

Answer:

1.) stability

2.) denser

3.) increase

4.) 1 millimeter

5.) regional

6.) more

7.) quartzite

8.) slate

Explanation:

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What mass in grams of hydrogen is produced by the reaction of 4.73 g of magnesium with 1.83 of water
Luba_88 [7]

The mass, in grams, of hydrogen that would be produced from the reaction will be 0.102 grams

<h3>Stoichiometric calculations</h3>

From the balanced equation of the reaction:

Mg + 2H_2O -- > Mg(OH)_2 + H_2

The mole ration of magnesium to water is 1:2.

Mole of 4.73 grams Mg = 4.73/24.3 = 0.195 moles

Mole of 1.83 grams water = 1.83/18 = 0.102 moles

Hence, water is limiting.

Mole ratio of water to the hydrogen produced = 2:1

Equivalent mole of hydrogen gas produced = 0.102/2 = 0.05 moles

Mass of 0.05 moles hydrogen gas = 0.05 x 2 = 0.102 grams

More on stoichiometric calculations can be found here: brainly.com/question/27287858

#SPJ1

4 0
2 years ago
Calculate the pH at the equivalence point when 22.0 mL of 0.200 M hydroxylamine, HONH2, is titrated with 0.15 M HCl. (Kb for HON
solniwko [45]

Answer:

pH = 3.513

Explanation:

Hello there!

In this case, since this titration is carried out via the following neutralization reaction:

HONH_2+HCl\rightarrow HONH_3^+Cl^-

We can see the 1:1 mole ratio of the acid to the base and also to the resulting acidic salt as it comes from the strong HCl and the weak hydroxylamine. Thus, we first compute the required volume of HCl as shown below:

V_{HCl}=\frac{22.0mL*0.200M}{0.15M}=29.3mL

Now, we can see that the moles of acid, base and acidic salt are all:

0.0220L*0.200mol/L=0.0044mol

And therefore the concentration of the salt at the equivalence point is:

[HONH_3^+Cl^-]=\frac{0.0044mol}{0.022L+0.0293L} =0.0858M

Next, for the calculation of the pH, we need to write the ionization of the weak part of the salt as it is able to form some hydroxylamine as it is the weak base:

HONH_3^++H_2O\rightleftharpoons H_3O^++HONH_2

Whereas the equilibrium expression is:

Ka=\frac{[H_3O^+][HONH_2]}{[HONH_3^+]}

Whereas Ka is computed by considering Kw and Kb of hydroxylamine:

Ka=\frac{Kw}{Kb}=\frac{1x10^{-14}}{9.10x10^{-9}}  \\\\Ka=1.10x10^{-6}

So we can write:

1.10x10^{-6}=\frac{x^2}{0.0858-x}

And neglect the x on bottom to obtain:

1.10x10^{-6}=\frac{x^2}{0.0858}\\\\x=\sqrt{1.10x10^{-6}*0.0858}=3.07x10^{-4}M

And since x=[H3O+] we obtain the following pH:

pH=-log(3.07x10^{-4})\\\\pH=3.513

Regards!

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Paraphin [41]
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Explanation:

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