All categories

3 years ago

our sun is considered average because its temperature and absolute magnitude place it in the __ range of the H-R diagram

Chemistry

1 answer:

Vlad [161]3 years ago

5 0

Answer:

middle

Explanation:

When the stars are plotted on the Hertzsprung-Russell diagram, most of them fall into the curve called the main sequence. The main sequence stars are 'average' because many stars fall into this category, and these stars burn hydrogen into helium in their core, and they are relatively young ( age of a few billion years) .

Our sun sits in the middle of the main sequence curve in the HR- diagram, which means the it is a yellow dwarf star with a surface temperature of about 6000 K, it has an average size, and it is hot enough for hydrogen fusion—in other words a typical star.

You might be interested in

PLEASE HELP! 25 POINTS!!!! I got the #1, just not #2 and #3. An industrial chemical company has opened a new plant that will pro

podryga [215]

Answer :

Part 1 : Balanced reaction, $3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

Part 2 : The theoretical yield of $NH_3$ gas = 440.96 g

Part 3 : The % yield of ammonia is 90.03 %

Solution : Given,

Mass of $N_2$ = 475 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $NH_3$ = 17 g/mole

Experimental yield of $NH_3$ = 397 g

Answer for Part (1) :

The balanced chemical reaction is,

$3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

Answer for Part (2) :

First we have to calculate the moles of $N_2$ .

$\text{Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molecular mass of }N_2}=\frac{475g}{28g/mole}=16.96moles$

From the given reaction, we conclude that

1 moles of $N_2$ gas react to give 2 moles of $NH_3$ gas

16.96 moles of $N_2$ gas react to give $\frac{2}{1}\times 16.96=33.92$ moles of $NH_3$ gas

Now we have to calculate the mass of $NH_3$ gas.

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

$\text{ Mass of }NH_3=(33.92moles)\times (17g/mole)=440.96g$

Therefore, the theoretical yield of $NH_3$ gas = 440.96 g

Answer for Part (3) :

Formula used for percent yield :

$\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theoretical yield of }NH_3}\times 100$

$\% \text{ yield of }NH_3=\frac{397g}{440.96g}\times 100=90.03\%$

Therefore, the % yield of ammonia is 90.03 %

3 0

3 years ago

A student makes a compound of sulfur and oxygen. She uses 5.00 g of sulfur and 4.99 g of oxygen, and all of the elemental substa

victus00 [196]

First, we apply the law of conservation of mass which states that the total mass in a system remains constant.
Therefore, there must be 5.00 g of sulfur and 4.99 g of oxygen in the product. Now, we determine the mass percentage using:

Mass % = (mass of sulfur x 100) / total mass of compound

Mass % = (5 * 100) / (5 + 4.99)

Mass % = 50.05%

The product contains 50.05% sulfur by mass.

7 0

3 years ago

Write the formulas of the following compounds:

dem82 [27]

Answer:

a) Li2CO3

b) NaCLO4

c) Ba(OH)2

d) (NH4)2CO3

e) H2SO4

f) Ca(CH3COO)2

g) Mg3(PO4)2

f) Na2SO3

Explanation:

a) 2Li + CO3 ↔ Li2CO3

b) NaOH * HCLO4 ↔ NaCLO4 + H2O

c) Ba + 2H2O ↔ Ba(OH)2 +

d) 2NH4 + H2CO3 ↔ (NH4)2CO3 + H2O

c) SO2 + NO2 +H2O ↔ H2SO4 + NOx

f) 2CH3COOH + CaO ↔ Ca(CH3COOH)2 + H2O

g) 3MgO + 2H3PO4 ↔ Mg3(PO4)2 + H2O

h) NaOH + H2SO3 ↔ Na2SO3 + H2O

6 0

3 years ago

What mass of salt (nacl) should you add to 1.48 l of water in an ice cream maker to make a solution that freezes at -13.4 ∘c ? a

blagie [28]

Answer is: mass of salt is 311,15 g.
V(H₂O) = 1,48 l · 1000 ml/l = 1480 ml.
m(H₂O) = 1480 g = 1,48 kg.
d(solution) = 1,00 g/ml.
ΔT(solution) = 13,4°C = 13,4 K.
Kf = 1,86 K·kg/mol; cryoscopic constant of water
i(NaCl) = 2; Van 't Hoff factor.
ΔT(solution) = Kf · b · i.
b(NaCl) = 13,4 K ÷ (1,86 K·kg/mol · 2).
b(NaCl) = 3,6 mol/kg.
n(NaCl) = 3,6 mol · 1,48 kg= 5,328 mol.
m(NaCl) = 5,328 mol · 58,4 g/mol = 311,15 g.

5 0

3 years ago

Read 2 more answers

Please help fast. I will give brainliest.

Sonbull [250]

Answer:39 min

Explanation:

5 0

2 years ago