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Wewaii [24]
3 years ago
10

Give a Convincing Argument that muscles are alive because they are made out of cells

Chemistry
1 answer:
lesantik [10]3 years ago
8 0
I think they are alive bc they are in a living body with cells
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Calculate the [H3O+] of solutions a and b; calculate the [OH-] solutions c and d.
patriot [66]

Answer:

I got the answers but it won't let me post it correctly on here....

Explanation:

9.) 10-2.76 =0.0174 [H30+]= 1.74*10-3 M

10.)10-3.65=0.00224  [H3O+] =2.24*10-2 M

11.)10-3.65=0.00224 [OH-]= 2.224*10-4M

12.)10-6.87=0.00000135  [OH-]= 1.35*10-7M

3 0
3 years ago
Common brass is a copper and zinc alloy containing 37.0% zinc by mass and having a density of 8.48g/cm3. A fitting composed of c
inysia [295]
First, we calculate the mass of the sample:

mass = density x volume
mass = 8.48 x 112.5
mass = 954 grams

Now, we will calculate the mass of each component using its percentage mass, then divide it by its atomic mass to find the moles and finally multiply the number of moles by the number of particles in a mole, that is, 6.02 x 10²³.

Zinc mass = 0.37 x 954
Zinc mass = 352.98 g
Zinc moles = 352.98 / 65
Zinc moles = 5.43
Zinc atoms = 5.43 x 6.02 x 10²³
Zinc atoms = 3.27 x 10²⁴

Copper mass = 0.63 x 954
Copper mass = 601.02 g
Copper moles = 601.02 / 64
Copper moles = 9.39
Copper atoms = 9.39 x 6.02 x 10²³
Copper atoms = 5.56 x 10²⁴
3 0
3 years ago
2. Calculate the mass of 3.47x1023 gold atoms.
lapo4ka [179]

3.47 x 10^{23} atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x 10^{23}

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X 10^{23}

from the relation,

1 mole of element contains 6.022 x 10^{23} atoms.

so no of moles of gold given = \frac{3.47 X 10^{23}  }{6.022 X 10^{23} }

0.57 moles of gold.

from the relation:

number of moles = \frac{mass}{atomic mass of 1 mole}

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x 10^{23} atoms of gold have mass of 113.44 grams

3 0
3 years ago
A 100.5 ml intraveneous (iv) solution contains 5.10 g glucose (c6h12o6). what is the molarity of this solution?
Pavlova-9 [17]
Answer is: <span>the molarity of this glucose solution is 0.278 M.
m</span>(C₆H₁₂O₆<span>) = 5.10 g.
n</span>(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆<span>) .
</span>n(C₆H₁₂O₆) = 5.10 g ÷ 180.156 g/mol.
n(C₆H₁₂O₆<span>) = 0.028 mol.
</span>V(solution) = 100.5 mL ÷ 1000 mL/L.
V(solution) = 0.1005 L.
c(C₆H₁₂O₆) = n(C₆H₁₂O₆) ÷ V(solution).
c(C₆H₁₂O₆) = 0.028 mol ÷ 0.1005 L.
c(C₆H₁₂O₆<span>) = 0.278 mol/L.</span> 
8 0
3 years ago
Pls help with this ​
Natalka [10]
Hirlkrrkkrkrmmrrmmejekkeowoow
3 0
3 years ago
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