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katrin2010 [14]
3 years ago
13

What the approximate mass of 24cm of silver, if the density is 10.5 g/cm

Chemistry
1 answer:
Alona [7]3 years ago
8 0
Use equation density = mass/volume
Known/Given

d= 10.5g/cm^3
v= 24.0 cm^3

Use that equation i just mentioned...
Solution
Rewrite the formula as the forumla for solving density into mass

Which is... mass = 10.5g/cm^3 × 24.0cm^3 = 252g

So thats the answer
252g

Hope i could help
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Could someone please tell show me steps to solve this, I am so confused.
jonny [76]

And this is y I should of paid attention in that one science class in middle school that taught this...

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6 0
3 years ago
Which solutions is more basic: a solution with a pH of 9 or a solution<br> with a pH of 12?
Nuetrik [128]

Answer:

Explanation:

pH= 7 is neutral, pH<7 - acidic solutions,

pH>7 - basic solution.

The more value of pH, the more basic solution is.

So, the solution with pH = 12 is more basic.

5 0
1 year ago
Calculate the mass percent (m/m) of a solution prepared by dissolving 45.09 g of NaCl in 174.9 g of H2O.
inysia [295]

Answer:

20.3 %  NaCl

Explanation:

Given data:

Mass of solute = 45.09 g

Mass of solvent = 174.9 g

Mass percent of solution = ?

Solution:

Mass of solution = 45.09 g + 174.9 g

Mass of solution = 220 g

The solute in 220 g is 45.09 g

220 g = 2.22 × 45.09

In 100 g solution amount of solute:

45.09 g/2.22 = 20.3 g

Thus m/m% = 20.3 %  NaCl

5 0
3 years ago
The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo
notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

6 0
3 years ago
What are the Sun's layers, beginning with the innermost and moving out.
Lemur [1.5K]
Core, radiative zone, convective zone, photosphere, chromosphere, corona 
4 0
3 years ago
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