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Saturns day is roughly 10 hours. it takes about 30 years for Saturn to return to the same point in space. How do saturns periods

sdas [7]

Earth 24 hours, 365 days, so c

6 0

3 years ago

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What would be the volume in millilitres of a blood sample of 2.15 microliters (ul)?

Zanzabum

2.15 x 10⁻³mL

Explanation:

Given parameter:

Volume of blood sample in uL = 2.15uL

Conversion uL → mL

micro- and milli- are both prefixes of sub-units.

liter is a unit of volume of a substance.

micro - is 10⁻⁶

milli- is of the order 10⁻³

The problem is converting from micro to milli:

if we multiply 10⁻⁶ by 10³ we would have our milli;

1000uL = 1mL

2.15uL : 2.15uL x $\frac{1mL}{1000uL}$ = 2.15 x 10⁻³mL

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Volume brainly.com/question/5055270

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4 0

3 years ago

According to the image, identify the number of neutrons in the most common isotope of aluminum.

wolverine [178]

Answer:

The number of neutron in the Aluminium Isotope is :

B. 14

Explanation:

Isotopes : These are the atoms which have same atomic number but have different mass number.

<u>This image shows the average atomic mass of Al element because it is in decimals</u>.

Atomic mass = 26.98154

(Note : mass number of single isotope can never be in decimals)

It is the average of mass of different isotopes of Al

Major Isotopes of $_{13}^{26.98154}\textrm{Al}$ are :

$_{13}^{26}\textrm{Al}$ ......atomic mass = 26
$_{13}^{27}\textrm{Al}$ .......atomic mass = 27

mass of Al given in image(26.98) is nearly equal to mass of 2nd isotope(27)

mass of $_{13}^{26.98154}\textrm{Al}\ \approx 27$

Now calculate the neutron in $_{13}^{27}\textrm{Al}$

Number of neutron = mass number - atomic number

= 27 - 13

Number of neutron = 14

(Atomic mass is same as mass number)

5 0

3 years ago

Calculate the enthalpy change, ∆H in kJ, for the reaction H2O(s) → H2(g) + 1/2O2(g). Use the following information: : +279.9 kJ

Len [333]

Answer:

+ 291.9 kJ

Solution:

The equation given is as;

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = ?

First, as we know the heat of formation of H₂O ₍l₎ is,

H₂ ₍g₎ + 1/2 O₂ ₍g₎ → H₂O ₍l₎ ΔH = - 285.9 kJ

Now, reversing the equation will reverse the sign of heat as,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

Also, we know that,

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

Now, adding last two equations,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

-----------------------------------------------------------------------------

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 291.9 kJ

5 0

3 years ago

The volume of 0.05 M H2SO4 is needed to completely neutralise 15ml of 0.1 M NaOH solution is

Feliz [49]

V(NaOH)=15 mL =0.015 L
C(NaOH)=0.1 mol/L
C(H₂SO₄)=0.05 mol/L

2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O

n(NaOH)=V(NaOH)C(NaOH)=2n(H₂SO₄)
n(H₂SO₄)=V(H₂SO₄)C(H₂SO₄)

V(NaOH)C(NaOH)=2V(H₂SO₄)C(H₂SO₄)

V(H₂SO₄)=V(NaOH)C(NaOH)/{2C(H₂SO₄)}

V(H₂SO₄)=0.015*0.1/{2*0.05}=0.015 L = 15 mL

5 0

3 years ago