Describe how this instrument works. Operation:

Chemistry

2 answers:

blsea [12.9K]3 years ago

4 0

Answ

Explanation:

that is like a satelite it helps detect signals from like outer space and things

icang [17]3 years ago

4 0

Answer:

An instrument takes the image that you are looking at and reflects it off the aluminum coating. The image then bounces off the coating to the glass. The glass is angle toward the eyepiece so that you have a full visual of the image you are trying to see. The telescope allows us to see further; they are able to collect and focus more light from distant objects than our eyes can alone. This is achieved by refracting or reflecting the light using lenses or mirrors.

Explanation:

Make me brainlist. :)

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What are families 3-12 on the periodic table called

fredd [130]

Answer:

transition elements

Explanation:

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How many grams of sodium are in 0.820 moles of na2so4?

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the molar mass of a compound is the sum of the products of the atomic masses by the number of atoms of the element.
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3 years ago

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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.

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