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3 years ago

6

Scientists us ____waves to study the characteristics of Earth Layers_ cannot travel through liquid, so the ______mus

t be liquid.

1 answer:

Lady bird [3.3K]3 years ago

6 0

Answer:

1)seismic 2) S-waves 3) outer core

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Read the given equation:

Assoli18 [71]

Answer:

D.

HCl is acid

Ca(OH)2 is base

CaCl2 is salt

5 0

3 years ago

Which is the correct name of a compound that is formed by Si and O? silicon dioxide silicon trioxygen disilicon dioxygen monosil

Flauer [41]

The correct is tricky, be careful. The right is silicon dioxyde (SiO2)

Silicon Oxides are written in the form SiOx, (0 <x <2), so:

there is no silicon trioxygen and disilicon dioxygen.

SiO is called silicon monoxide and not monosiicon oxygen, so this proposition is false.

All that remains is the silicon dioxide (SiO2) that is written correctly.

Silicon dioxide can be synthesized but also exists in abundance in nature. Silicon (Si) represents about 26% of the Earth's crust. Silica (SiO2), the natural form of silicon dioxide, accounts for about 60%.

8 0

3 years ago

Read 2 more answers

PLEASE HELP ME!!! NEED THE ANSWER FAST! WILL CHOOSE THE BRAINLIEST!!!

Vlad [161]

Remember that a cation will be smaller than its neutral atom, and an anion will be larger than its neutral atom. This would automatically eliminate answer choices A and D.

Also keep in mind that atomic radii decreases from left to right as you move along a periodic table. It also decreases from bottom up.
Atomic radii increases as you move from right to left and as you go from up to down.

As bromine is higher up in the periodic table than Iodine, it would have a smaller radius. Iodine would have a larger radius.

The correct answer is B. Br

8 0

3 years ago

Along with subduction, what other process changes sedimentary rocks to metamorphic rocks?

muminat

It would be weathering because of all the heat and pressure.

7 0

3 years ago

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

5 0

3 years ago