Which molecule make up proteins?

"Two major varieties of igneous rock are _______ and ________. What is the difference between these two types of igneous rock?"

grandymaker [24]

Answer:

Intrusive and Extrusive igneous rocks.

Explanation:

Igneous rocks are defined as those rocks that are formed when magma undergoes the process of crystallization and solidification at or below the earth's surface. For example, Granite, Rhyolite, Gabbro and Diorite.

The igneous rocks are of two different types, namely-

Intrusive igneous rocks- This type of igneous rocks are formed when the magma crystallizes below or within the earth's crust. For example, Granite.

Extrusive igneous rocks- This type of igneous rocks are formed when the magma crystallizes and solidifies at the surface of the earth. For example, Basalt.

8 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

marin [14]

Answer:

$\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}$

Explanation:

1. Calculate the moles of copper(II) hydroxide

$\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}$

2. Calculate the molecules of copper(II) hydroxide

$\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}$

6 0

3 years ago

Methods scientists use to determine if a change was physical or chemical.

Tpy6a [65]

Chemical reactions can be identified when there is a change in color, energy is produced, change in odor, or if new substance forms.

7 0

4 years ago

Read 2 more answers

What evidence supports a conservation law?

True [87]

Answer:

The answer is B.

Explanation:

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations.

According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

The law of conservation of mass is useful for a number of calculations and can be used to solve for unknown masses, such the amount of gas consumed or produced during a reaction.

3 0

3 years ago

What is the freezing point of a solution of 465 g of sucrose c12h22o11 dissolved in 575 ml of water?

Amanda [17]

Answer:

The freezing point of the solution is - 4.39 °C.

Explanation:

We can solve this problem using the relation:

ΔTf = (Kf)(m),

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

density of water = 1 g/mL.

So, the mass of 575 mL is 575 g = 0.575 kg.

m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.

∴ ΔTf = (Kf)(m) = (-1.86 °C/m)(2.36 m) = - 4.39 °C.

∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.

∴ The freezing point of the solution is - 4.39 °C.

6 0

3 years ago