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3 years ago

10

I need help I’ll give you brainliest

2 answers:

FinnZ [79.3K]3 years ago

6 0

Pretty sure it’s 3 otherwise idk

Svetlanka [38]3 years ago

5 0

Answer:

2 or 3 i think

Explanation:

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All atomic nuclei, except those of ordinary hydrogen, contain neutrons. <br> a. True<br> b. False

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This would be false

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3 years ago

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The strength of the gravitational force between two objects depends on __________.

irga5000 [103]

Should be their masses. Because t<span>he strength of the gravitational force between two objects depends on two factors, mass and </span>distance<span>. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.</span>

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3 years ago

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Interstellar space has an average temperature of about 10 K , and an average density of hydrogen atoms of about one hydrogen ato

Bas_tet [7]

Answer:

$2.2508\times 10^{19} m$ meter the mean free path of hydrogen atoms in interstellar space.Explanation:

The mean free path equation is given as:

$\lambda =\frac{1}{\sqrt{2}\pi d^2 n}$

Where"

d = diameter of hydrogen atom in meters

n = number of molecules per unit volume

We are given: d = 100 pm = $100\times 10^{-12}= 10^{-10} m$

$n = 1 m^{-3}$

$\lambda =\frac{1}{\sqrt{2}\pi (10^{-10} m)^2\times 1 m^{-3}}$

$\lambda =2.2508\times 10^{19} m$

$2.2508\times 10^{19} m$ meter the mean free path of hydrogen atoms in interstellar space.

7 0

3 years ago

Gold has a density of 1,200 lb./ft. What is the density of gold in g/em? For conversion factors use I lb. 453.6 g, and l inch-2.

Jlenok [28]

<u>Answer:</u> The density of gold in $g/cm^3$ is $19.22g/cm^3$

<u>Explanation:</u>

Density is defined as the ratio of mass of the object and volume of the object. Mathematically,

$\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}$

We are given:

Density of gold = $1200lb/ft^3$

Using conversion factors:

1 lb = 453.6 g

1 feet = 12 inches

1 inch = 2.54 cm

Converting given quantity into $g/cm^3$ , we get:

$\Rightarrow (\frac{1200lb}{ft^3})\times (\frac{453.6g}{1lb})\times (\frac{1ft}{12inch})^3\times (\frac{1inch}{2.54cm})^3\\\\\Rightarrow 19.22g/cm^3$

Hence, the density of gold in $g/cm^3$ is $19.22g/cm^3$

6 0

3 years ago

Addition of an excess of lead (II) nitrate to a 50.0mL solution of magnesium chloride caused a formation of 7.35g of lead (II) c

KiRa [710]

Answer:

[Cl⁻] = 0.016M

Explanation:

First of all, we determine the reaction:

Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓ + Mg(NO₃)₂(aq)

This is a solubility equilibrium, where you have a precipitate formed, lead(II) chloride. This salt can be dissociated as:

PbCl₂(s) ⇄ Pb²⁺ (aq) + 2Cl⁻ (aq) Kps

Initial x

React s

Eq x - s s 2s

As this is an equilibrium, the Kps works as the constant (Solubility product):

Kps = s . (2s)²

Kps = 4s³ = 1.7ₓ10⁻⁵

4s³ = 1.7ₓ10⁻⁵

s = ∛(1.7ₓ10⁻⁵ . 1/4)

s = 0.016 M

3 0

3 years ago