All categories

3 years ago

Please explain how we judge the brightness of stars.

1 answer:

6 0

Astronomers can measure it by way of luminosity which is the power of a star or the amount of energy (light) the star admits from its surface. they also measure the brightness of a start as if it were to appear 32.6 light years from Earth

You might be interested in

A balloon contains 0.5 L of air at 101.5 kPa. You squeeze it to a volume of 0.25 L. What is the pressure of air inside the ballo

Igoryamba

The answer would be D. 203kPa

4 0

3 years ago

What is the most mass of (NH4)2CO3 (H=1,C=12,N=14,O=16

Agata [3.3K]

Answer:

96.09 g/mol

Explanation:

You just need to first get the atomic weights of the elements involved. You can easily get these from your periodic table.

If you are going to do this properly, please use the weight with at least two decimal places for accuracy (e.g. 15.99 g/mol).

Also, please take note that I will be using the unit g/mol for all the weights. Thus,

Step 1

N = 14.01 g/mol

H = 1.008 g/mol

O = 16.00 g/mol

C = 12.01 g/mol

Since your compound is

(

)

, you need to multiply the atomic weights by their subscripts. Therefore,

Step 2

N = 14.01 g/mol × 2 =

28.02 g/mol

H = 1.008 g/mol × (4×2) =

8.064 g/mol

O = 16.00 g/mol × 3 =

48.00 g/mol

C = 12.01 g/mol × 1 =

12.00 g/mol

To get the mass of the substance, we need to add all the weights from Step 2.

Step 3

molar mass of

(

)

(28.02 + 8.064 + 48.00 + 12.01) g/mol

96.09 g/mol

this is a google search and a example i hope is helps to solve

6 0

3 years ago

In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.

trasher [3.6K]

Answer:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction

The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction

The second run has twice the surface area - yes, 44 sqcm to 22 sqcm

Explanation:

A catalyst is a material which speeds up a reaction without being consumed in the process. A heterogeneous catalyst is one which is of a different phase than the reactants. The effectiveness of a catalyst is dependent on the available surface area. The first step for this question is to determine the total available surface area of catalyst in both processes.

Step 1: Determine radius of large sphere

$V_{L}=\frac{4}{3}*pi*{r_{L}}^{3}$

$10=\frac{4}{3}*pi*{r_{L}}^{3}$

${r_{L}}^{3}=\frac{7.5}{pi}$

$r_{L}=1.337cm$

Step 2: Determine surface area of large sphere

$A_{L}=4*pi*{r_{L}}^{2}$

$A_{L}=4*pi*1.337^{2}$

$A_{L}=22.463 cm^{2}$

Step 3: Determine radius of small sphere

$V_{s}=\frac{4}{3}*pi*{r_{s}}^{3}$

$1.25=\frac{4}{3}*pi*{r_{s}}^{3}$

${r_{s}}^{3}=\frac{0.9375}{pi}$

$r_{s}=0.668cm$

Step 4: Determine surface area of small sphere

$A_{s}=4*pi*{r_{s}}^{2}$

$A_{s}=4*pi*0.668^{2}$

$A_{s}=5.607 cm^{2}$

Step 5: Determine total surface area of 8 small spheres

$A_{S}=8*A_{s}$

$A_{S}=8*5.607$

$A_{S}=44.856 cm^{2}$

$A_{L}=22.463 cm^{2}$ - Surface area of 1 large sphere

$A_{S}=44.856 cm^{2}$ - Surface area of 8 small spheres

Options:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction
The second run will be slower - false, the increased surface area of catalyst will increase the rate of reaction
The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction
The second run has twice the surface area - yes, 44 sqcm to 22 sqcm
The second run has eight times the surface area - no, 44 sqcm to 22 sqcm
The second run has 10 times the surface area - no, 44 sqcm to 22 sqcm

7 0

3 years ago

Read 2 more answers

To what volume should 25ml of 15m nitric acid be diluted to prepare a 3m solution

Maurinko [17]

We can use the dilution formula to find the volume of the diluted solution to be prepared

c1v1 = c2v2

Where c1 is concentration and v1 is volume of the concentrated solution

And c2 is concentration and v2 is volume of the diluted solution to be prepared

Substituting the values in the equation

15 M x 25 mL = 3 M x v2

v2 = 125 mL

The 25 mL concentrated solution should be diluted with distilled water upto 125 mL to make a 3 M solution

6 0

4 years ago

An individual is hospitalized and the initial blood work indicates high levels of hco3- in the blood and a ph of 7. 47. This wou

expeople1 [14]

An individual is hospitalized and the initial blood work indicates high levels of $HCO_{3} ^{-}$ in the blood and a pH of 7. 47. This would indicate the individual probably has compensated respiratory acidosis.

A chronic illness usually leads to compensated respiratory acidosis because the kidneys have time to adjust to the delayed onset. Even if the $PCO_{2}$ is elevated in a compensated respiratory acidosis, the pH is within the usual range.

The kidneys counteract a respiratory acidosis by increasing the amount of $HCO_{3}$ that tubular cells reabsorb from the tubular fluid, the amount of $H^{+}$ that collecting duct cells secrete while also producing $HCO_{3}$ , and the amount of $NH_{3}$ buffer that is formed through ammoniagenesis.

Respiratory acidosis is frequently brought on by hypoventilation as a result of: breathing depression , paralysis of the respiratory muscles, diseases of the chest wall , abnormalities of the lung parenchyma and abdominal squeezing.

Learn more about Respiratory acidosis here;

brainly.com/question/9694207

#SPJ4

5 0

1 year ago