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Leokris [45]
3 years ago
15

The blackbody curve for a star named Gamma shown below. What is the surface temperature of this star rounded to the nearest whol

e number? 9,666 K
Chemistry
1 answer:
Natali5045456 [20]3 years ago
3 0

Answer:

yes

Explanation:

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3. Determine the moles of sodium, Na, containing 7.9x1024 atoms.​
-BARSIC- [3]

Answer:

12.7mol Na.

Explanation:

Hello there!

In this case, according to the concept of mole, which stands for the amount of substance, we can recall the concept of Avogadro's number whereby we understand that one mole of any substance contains 6.022x10²³ particles, for the given atoms of sodium, we can calculate the moles as shown below:

7.9x10^{23}atoms*\frac{1mol}{6.022x10^{23}atoms} \\\\

Thus, by performing the division we obtain:

12.7molNa

Regards!

8 0
3 years ago
Circle the correct one from given alternatives.
mash [69]

Answer:

24 is the correct anwer

this the anwer text this u no

3 0
2 years ago
Read 2 more answers
To a 0.0001 m solution of mg(no3)2, naoh was added to a final concentration of 0.001m did a precipitate form?
Natalija [7]

I looked on a solubility chart to answer this question, and hydroxides are generally insoluble (with some exceptions of course). However, it says to consider Mg(OH)_{2} as an insoluble substance, though it may be moderately soluble.


The answer that you are most likely looking for is: Yes, a precipitate does form - this is due to the double placement reaction:


Mg(NO_{3})_{2}_{(aq)} + 2NaOH_{(aq)} → Mg(OH)_{2} {(s)} + 2NaNO_{3}_{(aq)}

8 0
3 years ago
Suppose that a certain biologically important reaction is quite slow at physiological temperature (37 oC) in the absence of a ca
Oksi-84 [34.3K]

Answer:

30 kJ

Explanation:

Arrhenius equation is given by:

k=Aexp(-Ea/RT)\\

Here, k is rate constant, A is Pre-exponential factor, Ea is activation energy and T is temperature.

taking natural log of both side

ln k = ln A - Ea/RT

In Arrhenius equation, A, R and T are constant.

Therefore,

ln\frac{k_2}{k_1} =\frac{Ea_1-Ea_2}{RT}

Ea_1-Ea_2 is the  lowering in activation energy by enzyme,

R = 8.314 J/mol.K

T = 37°C + 273.15 = 310 K

\frac{k_2}{k_1} =1\times 10^5

ln 1\times 10^5 =\frac{Ea_1-Ea_2}{RT}\\{Ea_1-Ea_2} = 11.512 \times 8.314 \times 310\\=29670\ J\\=30\ kJ

4 0
3 years ago
(C6H6) can be biodegraded by microorganism. if 30 mg of benzene is present, what amount of oxygen required for biodegradation, n
quester [9]

Answer:

36.92 mg of oxygen required for bio-degradation.

Explanation:

5C_6H_6+15O_2\rightarrow 12CO_2+15H_2O

Mass of benzene = 30 mg = 0.03 g (1000 mg = 1 g )

Moles benzene =\frac{0.03 g}{78 g/mol}=0.0003846 mol

According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.

Then 0.0003846 mol of benzene will react with:

\frac{15}{5}\times 0.0003846 mol=0.0011538 mol of oxygen gas

Mass of 0.0011538 moles of oxygen gas:

0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg

36.92 mg of oxygen required for bio-degradation.

4 0
3 years ago
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