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2 years ago

What laws do astronomers use to indirectly learn about stars?

Chemistry

1 answer:

Maru [420]2 years ago

6 0

Answer

Laws that are used by astronomers to learn about stars indirectly are as given below:-

Kirchhoff's First Law
Cassini's laws

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Commercially available aqueous nitric acid has a density of 1.42 g/mL and a concentration of 16 M. Calculate the mass percent of

luda_lava [24]

Answer:

The mass % of HNO3 in the solution is 71.0 %

Explanation:

Step 1: Data given

Density of HNO3 = 1.42 g/mL

Concentration = 16 M = 16 mol /L

Molar mass HNO3 = 63.01 g/mol

Assume the volume = 1L or 1000 mL

Step 2: Calculate mass of the solution

Mass = density * volume

Mass = 1.42 g/mL * 1000 mL

Mass = 1420 grams

Step 3: Calculate moles HNO3

Moles = molarity * volume

Moles = 16 M * 1L

Moles = 16 moles

Step 4: Calculate mass HNO3

Mass HNO3 = moles * molar mass

Mass HNO3 = 16.0 moles * 63.01 g/mol

Mass HNO3 = 1008.16 grams

Step 5: Calculate the mass percent

mass % = (1008.16 grams / 1420 grams) *100%

mass % = 71.0 %

The mass % of HNO3 in the solution is 71.0 %

5 0

3 years ago

Please anyone help me

sweet-ann [11.9K]

Answer:

F. a dark solid formed on the zinc.

-Is because when the solid is formed it is called the "precipitate", which is commonly know as chemical reaction.

8 0

3 years ago

A chemist wishing to do an experiment requiring 47Ca2+(half-life = 4.536 days) needs 5.00 g of the nuclide. What mass of 47CaCO

Artyom0805 [142]

Answer: The answer is 6.78 grams.

Explanation: The equation used for solving this type of problems is:

$\frac{N}{N_0}=(\frac{1}{2})^n$

where, $N_0$ is the initial amount of radioactive substance, N is the remaining amount and n is the number of half lives.

Number of half lives is calculated on dividing the given time by the half life.

n = time/half life

Time is given as 48.0 hours and the half life is given as 4.536 days. let's make the units same and for this let's convert the half life from days to hours.

$4.536days(\frac{24hours}{1day})$

= 108.864 hours

So, $n=\frac{48.0}{108.864}$ = 0.441

Since 5.00 g is the required amount when the radioactive substance is delivered to the scientist, it would be the final amount that is N. We need to calculate the initial amount. Let's plug in the values in the equation:

$\frac{5.00}{N_0}=(\frac{1}{2})^0^.^4^4^1$

$\frac{5.00g}{N_0}=0.737$

$N_0=\frac{5.00g}{0.737}$

$N_0$ = 6.78 g

So, 6.78 g of the radioactive substance needs to be ordered.

4 0

4 years ago

If m = 45g and V = 15ml, what is D in g/ml?

ch4aika [34]

Answer:

3 g/mL

Explanation:

We know that the density of an object can be measured by dividing its mass (g) to its volume (mL).

Formula

D=m/v

Given data:

Mass= 45 g

Volume= 15 mL

Now we will put the values in formula:

D=45 g/ 15 mL= 3 g/mL

6 0

4 years ago

Using the following portion of the activity series for oxidation half-reactions, determine which combination of reactants will r

dusya [7]

The question is incomplete, the complete question is:

Using the following portion of the activity series for oxidation half-reactions, determine which combination of reactants will result in a reaction.

$Li(s)\rightarrow Li^+(aq)+e^-$

$Al(s)\rightarrow Al^{3+}(aq)+3e^-$

A) Li(s) with Al(s)

B) Li(s) with $Al^{3+}$ (aq)

C) $Li^+$ (aq) with Al(s)

D) $Li^+$ (aq) with $Al^{3+}$ (aq)

Answer: The correct option is B): Li(s) with $Al^{3+}$ (aq)

Explanation:

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction.

The chemical species will undergo a reduction reaction if the value of standard reduction potential is more positive or less negative.

For the given half-reactions:

$Li(s)\rightarrow Li^+(aq)+e^-;E^o_{Li^+/Li}=-3.04V$

$Al(s)\rightarrow Al^{3+}(aq)+3e^-;E^o_{Al^{3+}/Al}=-1.662V$

As the value of standard reduction potential of aluminium is less negative. Thus, it undergoes reduction reaction and lithium will undergo oxidation reaction.

The half-reaction follows:

Oxidation half-reaction: $Li(s)\rightarrow Li^+(aq)+e^-$ ( × 3)

Reduction half-reaction: $Al^{3+}(aq)+3e^-\rightarrow Al(s)$

Overall cell-reaction: $3Li(s)+Al^{3+}(aq)\rightarrow 3Li^+(aq)+Al(s)$

Hence, the correct option is B): Li(s) with $Al^{3+}$ (aq)

5 0

3 years ago

What laws do astronomers use to indirectly learn about stars?​

What laws do astronomers use to indirectly learn about stars?