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3 years ago

14

3. Sketch a simple model that shows why the constellations change over the course of a year. Then, annotate the model to explain

why you see different stars throughout the year.

1 answer:

Lorico [155]3 years ago

8 0

Constellation and i will not be in a new

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HELPP ME PLEASE!! What would be the density of a 40 g object that displaces 10 ml of water?

Snezhnost [94]

Answer:

<h2>The answer is 4.0 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 40 g

volume = 10 mL

The density of the object is

$density = \frac{40}{10} = 4 \\$

We have the final answer as

<h3>4.0 g/mL</h3>

Hope this helps you

5 0

3 years ago

If one were asked to choose between paper or plastic which would be chosen and which one would benefit the environment?

vekshin1

It would be plastic Becaus we can recreate things with plastic and other things that are recyclable. Since,we mostly cut down trees everyday for paper,not knowing that we are wasting natural resources, and destroying animal homes and food.

6 0

4 years ago

Correct answer please i need it before 12:07

choli [55]

Answer:

xenon

Explanation:

Xenon is a non-metal, odorless gas, consist of single atoms, colorless, and 8 valence electrons out of everything else on this list.

6 0

2 years ago

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

7 0

2 years ago

In the reaction a b c, doubling the concentration of a doubles the reaction rate and doubling the concentration of b does not af

PolarNik [594]

The reaction must be a + b --> c

Then you can predict a reaction rate, r o the type r = k * a^n * b^m

Given that the reaction rate is not affected by the concentration of b you can state that m = 0 and r = k * a^n.

Now given, that there is a proportional relation between the reaction rate and a (double a gives double rate), then n = 1 and r = k*a. You can verify that if you dobule a r also doubles.

Answer: r = k*a

3 0

3 years ago