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LenaWriter [7]
2 years ago
8

C1 Progress quiz: Atomic structure 2 - test

Chemistry
1 answer:
myrzilka [38]2 years ago
5 0

Answer:

0.11 nm

Explanation:

1.1 x 10-10 m

The goal is to convert the atomic radius from meters (m) to nanometres (nm). We do this by multiplying the value by 10^9.

This is given as;

1.1 * 10^{-10}  * 10^{9}\\1.1 * 10^{-10 + 9}\\1.1 * 10^{-1}\\0.11

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If 0.507J of heat leads to a 0.007 degree C change in water, what mass is present?
Mamont248 [21]

Answer:

17.3 g

Explanation:

<u>Given the following data;</u>

  • Quantity of heat, Q = 0.507 J
  • Temperature = 0.007°C
  • Specific heat capacity of water = 4.2 J/g°C

Mathematically, Heat capacity is given by the formula;

Q = MCT

Where;

  • Q represents the heat capacity or quantity of heat.
  • M represents the mass of an object.
  • C represents the specific heat capacity of water.
  • T represents the temperature.

Making "M" the subject of formula, we have;

M = \frac {Q}{CT}

Substituting the values into the formula, we have;

M = \frac {0.507}{4.2*0.007}

M = \frac {0.507}{0.0294}

<em>Mass, m = 17.3 grams</em>

8 0
2 years ago
A large highway barrier is 1 meter wide, by 1 meter tall, by 2 meters long.
Romashka [77]

Answer:

when mass is  1×10⁴ Kg then density is 5 g/cm³.

when mass is 104 Kg then density is 5.2 × 10⁻² g/ cm³.

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Given data:

mass = 1×10⁴ Kg

volume= w ×l× h = 1×2× 1 = 2 m³

density = ?

first of all we will convert the given volume meter cube to cm³:

we know that  

2×1000000 = 2 × 10⁶ cm³

Now we will convert the mass into gram.

1 Kg = 1000 g

1×10⁴ × 1000 = 1 ×10⁷ g

Now we will put the values in the formula,

d = m/v

d = 1 ×10⁷ g / 2×10⁶ cm³

d = 0.5 × 10¹ g/cm³

   or

d = 5 g/cm³

If mas is 104 Kg:

104 × 1000 = 104000 g

d= m/v

d = 104000 g / 2×10⁶ cm³

d= 52000 ×10⁻⁶ g/ cm³

d= 5.2 × 10⁻² g/ cm³

7 0
3 years ago
Ethanol, C2H6O, is most often blended with gasoline - usually as a 10 percent mix - to create a fuel called gasohol. Ethanol is
weeeeeb [17]

Answer:

This means 463 grams of ethanol would provide less amount of energy

Explanation:

Step 1: Data given

Heat of combustion of ethanol = 326.7 kcal/mol

The heat of combustion of octane =  1.308*10³ kcal/mol

Mass of octane = 463 grams

Molar mass octane = 114.23 g/mol

Molar mass ethanol = 46.07 g/mol

Step 2: Calculate moles octane

Moles octane = mass octane / molar mass octane

Moles octane = 463 grams / 114.23 g/mol

Moles octane = 4.05 moles

Step 3: Calculate energy of combustion of 4.05 moles octane

Combustion of 1 mol octane gives us: 1.308 * 10³ kcal/mol

Combustion of 4.05 moles octane gives us 4.05 * 1.308 * 10³ kcal/mol = <u>5.30 * 10³ kcal</u>

This means the combustion reaction of 463 grams of octane gives us 5.30 * 10³ kcal

Step 4:

Heat of combustion of ethanol = 326.7 kcal/mol

OR in words: combustion of 1 mol ethanol gives us 326.7 kcal energy

Moles ethanol = 463 grams / 46.07 g/mol

Moles ethanol = 10.05 moles

Since combustion of 1 mol ethanol gives us 326.7 kcal

10.05 moles ethanol will give us = 10.05 * 326.7 = 3283.3 kcal = <u>3.28 * 10³ kcal</u>

<u />

5.30 * 10³ kcal > 3.28 * 10³ kcal

This means 463 grams of ethanol would provide less amount of energy

3 0
3 years ago
Calculate the solubility of mn(oh)2 in grams per liter when buffered at ph=7.0. assume that buffer capacity is not exhausted
Vanyuwa [196]
When PH + POH = 14 
∴ POH = 14 -7 = 7

when POH = -㏒[OH-]

          7    = -㏒ [OH-]
∴[OH-] = 10^-7

by using ICE table:

           Mn(OH)2(s) ⇄  Mn2+ (aq)  + 2OH-(aq)
initial                              0                     10^-7
change                           +X                      +2X
Equ                                 X                  (10^-7 + 2X)

when Ksp = [Mn2+][OH-]^2

when Ksp of Mn(OH)2 = 4.6 x 10^-14

by substitution:

4.6 x 10^-14 = X*(10^-7+2X)^2  by solving this equation for X

∴ X =2.3 x 10-5 M

∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol

= 0.002 g/ L
3 0
3 years ago
Read 2 more answers
Calculate the mass percent composition of sulfur in Al2(SO4)3.
Anastaziya [24]
The molar mass of aluminum sulftae is 342.14 g/mol.

Since the subscript shows that there are 3 sulfurs within the substance, the total mass of sulfur is 96.21g/mol

Now take the mass of the sulfur and divide it by the molar mass of aluminum sulfate, then multiply by 100:
(96.21/342.15)(100) = 28.1% mass composition of sulfate
6 0
3 years ago
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