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Radda [10]
3 years ago
6

The radius of an atom of element X is 1.2x10m The radius of the centre of the atom is 1/10000 the radius of the atom Calculate t

he radius of the centre of an atom of element X
Chemistry
1 answer:
Alika [10]3 years ago
8 0

Answer:

radius=1.2x10^{-3}m

Explanation:

Hello,

Based on the given information, the radius is computed via:

radius=1.2x10m*\frac{1}{10000}=1.2x10^{-3}m

By assuming that the radius of the element is 1.2x10m, nevertheless, that is a pretty much big radius for an atom in meters, maybe a power is missing, but you can modify it by just including it in the aforesaid formula.

Best regards.

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The force produced by the pull of the earth's gravity on an object is called what?
umka2103 [35]
The force is called weight
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When a physical change occurs what cannot be affected as well?
vitfil [10]

Answer:

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Explanation:

7 0
3 years ago
You are given 50.0 ml of 1.50 m hno3. how many ml of 0.81 m naoh are needed to neutralize it?
svet-max [94.6K]
V(HNO₃) = 50.0 mL in liters = 50.0 /1000 =0.05 L
M(HNO₃) = 1.50 M

Number of moles HNO₃ :

n = M x V

n = 1.50 x 0.05

n = 0.075 moles of HNO₃

HNO₃ + NaOH = H₂O + NaNO₃

1 mole HNO₃ -------- ---1 mole NaOH
0.075 moles HNO₃ ---- ?

moles NaOH = 0.075 * 1 / 1

 = 0.075 moles of NaOH

V ( NaOH ) :

M = n / V

0.81 = 0.075 / V

V = 0.075 / 0.81

V =<span> 0.0925 L or  92.5 mL </span>

<span>hope this helps!</span>
7 0
3 years ago
The formula for a compound of Li+ ions and Br- ions is written LiBr. Why can’t it be written Li2Br? Why isn’t it written BrLi?
maksim [4K]

Lithium has charge of +1 and bromide has charge of - 1. So they combine to form the compound lithium bromide which is expressed as LiBr.

<h3><u>Explanation:</u></h3>

Lithium is an alkali metal placed in group 1 or periodic table. It has a valency of 1 which is achieved as lithium loses an electron to achieve a charge of +1.

Bromine is a halogen which is placed in group 17 of periodic table. It has a valency of 1 which is achieved as bromine looses an election to achieve a charge of - 1.

Lithium is the cation and bromide is the anion. So lithium is written in front and bromine following the cation. And as both of their valencies are 1, so they form the compound LiBr.

3 0
3 years ago
John dissolves .5g of a white powder in 25g of benzene (FP 5oC) (kf benzene is 5.1) and finds the solution freezes at 3.7oC. Det
navik [9.2K]

Answer:

The compound has a molar mass of 78.4 g/mol

Explanation:

Step 1: data given

Mass of a sample = 0.5 grams

Mass of benzene = 25 grams

Freezing poing = 5 °C

Kf of benzene = 5.1 °C/m

Freezing point solution = 3.7 °C

Step 2: Calculate molality

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 5.0 - 3.7 = 1.3 °C

⇒with i = the can't hoff factor = 1

⇒with Kf = the freezing point depression constant of benzene = 5.1 °C/m

⇒with m = the molality

1.3 = 5.1 * m

m = 1.3 / 5.1

m = 0.255 moles /kg

Step 3: Calculate moles

Molality = moles / mass benzene

0.255 molal = moles / 0.025 kg

Moles = 0.255 molal * 0.025 kg

Moles = 0.006375 moles

Step 4: Calculate molar mass of the compound

Molar mass compund = mass / moles

Molar mass compound = 0.5 grams / 0.006375 moles

Molar mass compound = 78.4 g/mol

The compound has a molar mass of 78.4 g/mol

7 0
3 years ago
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