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Hitman42 [59]
3 years ago
11

A rock that forms from many small fragments of other rocks is a what type of rock

Chemistry
1 answer:
neonofarm [45]3 years ago
6 0
The answer would be Sedimentary
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How are atoms structured
Vladimir79 [104]

Answer:

they got protons, electrons, and neutrons

Explanation:

4 0
3 years ago
Read 2 more answers
The empirical formula of a compound is determined to be CH2O, and its molecular mass is found to be 90.087 g/mol. Determine the
enot [183]
The empirical formula of a compound is determined to be CH2O, and its molecular mass is found to be 90.087 g/mol. Determine the molecular formula of the compound, showing your solution.

Answer: This is actually quite simple, first we have to calculate the molar mass of empirical unit. Therefore we have 12+2*1+16 = 30. Then we solve 90/30 = 3. Finally we end up with 3*(CH2O) --> C3H6O3.

I hope it helps, Regards.
5 0
3 years ago
Read 2 more answers
If you start with 64g of a radioactive element how many half-lives would occur before 8g remain?
Nonamiya [84]

Answer:

3 half-lives

Explanation:

The half-life is the time that it takes to a radioactive element to decay to half of its initial amount.

Let's suppose we start with 64 g of the radioactive element.

  • After 1 half-life, the mass of the element will be 32 g.
  • After 2 half-lives, the mass of the element will be 16 g.
  • After 3 half-lives, the mass of the element will be 8 g.
3 0
3 years ago
Consider the reaction:
stich3 [128]

Answer:

\large \boxed{\text{-851.4 kJ/mol}}

Explanation:

2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s); ΔᵣH = ?

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})

                            2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s)

ΔfH°/kJ·mol⁻¹:         0         -824.3         -1675.7         0

\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [1(-1675.7) + 2(0)] - [2(0) - 1(-824.3)]\\& = & -1675.7 + 824.3\\& = & \textbf{-851.4 kJ/mol}\\\end{array}\\\text{The enthalpy change is } \large \boxed{\textbf{-851.4 kJ/mol}}

7 0
4 years ago
Find the density of the object below,
Kitty [74]

Answer:

19.32g/cm³

Explanation:

Density = m/w

Density = 10g/0.5176cm³

Density = 19.32 g/cm³

4 0
3 years ago
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