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VikaD [51]
3 years ago
5

a rock has a density of 15 g/cm3 and a mass of 50 g is dropped into a 100ml graduated cylinder containing 55 ml of water. to wha

t height will the water rise in the cylinder
Chemistry
2 answers:
WARRIOR [948]3 years ago
7 0
Answer is: the height of cylinder is 58.33 ml.
d(rock) = 15 g/cm³.
m(rock) = 50 g.
V(rock) = m(rock) ÷ d(rock).
V(rock) = 50 g ÷ 15 g/cm³.
V(rock) = 3.33 cm³ = 3.33 ml..
h(<span>cylinder) = V(rock) + V(water).
</span>h(cylinder) = 3.33 cm³ + 55 cm³.
h(cylinder) = 58.33 cm³.
fomenos3 years ago
4 0
We know that, Density = Mass/Volume


Therefore, Volume = Mass /Density = 50/15 = 3.33 ml

Thus, volume occupied by rock = 3.33 ml
 
Thus, total volume occupied in <span>graduated cylinder = 55 ml + 3.33 = 58.33 ml 
</span>
∴volume of <span>graduated cylinder will rise by 3.33 cm3.
</span>
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Answer:

6.02 x 10²³ electrons

Explanation:

Given parameters:

Mass of H₂  = 2g

Unknown:

Number of electrons = ?

Solution:

To find the number of electrons, we must determine the number of moles of H₂ first.

 Number of moles  = \frac{mass}{molar mass}  

  Molar mass of H₂   = 2(1)  = 2g/mol

Number of moles  = \frac{2}{2}   = 1mol  

1 mole of a substance contains 6.02 x 10²³ particles

the particles can be protons, neutrons, electrons

So,

  2g of H₂ will contain 6.02 x 10²³ electrons.

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4 0
1 year ago
A container of carbon dioxide has a volume of 260 cm at a temperature of 22.0°C. If the
Oxana [17]

Answer:

V₂ =279.4 cm³

Explanation:

Given data:

Initial volume = 260 cm³

Initial temperature = 22.0°C

Final temperature = 44.0°C

Final volume = ?

Solution;

22.0°C (22+ 273 = 295k)

44.0°C(44+273 = 317k)

Formula:

According to Charles's law

V₁/T₁ = V₂/T₂

Now we will put the values in formula:

V₂ = V₁×T₂ / T₁

V₂ = 260 cm³ × 317k / 295k

V₂ = 82420 cm³. k  / 295k

V₂ =279.4 cm³

3 0
4 years ago
At what step of glycolysis one NADH+ H+ is formed?
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Answer:

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3 0
3 years ago
The protein catalase catalyzes the reaction The Malcolm Baldrige National Quality Award aims to:
Marina CMI [18]

The question is missing a part, so the complete question is as follows:

The protein catalase catalyzes the reaction The Malcolm Bladrigde National Quality Awards aims to: 2H2O2 (aq) ⟶ 2H2O (l) + O2 (g) and has a Michaelis-Menten constant of KM = 25mM and a turnover number of 4.0 × 10 7 s -1. The total enzyme concentration is 0.012 μM and the intial substrate concentration is 5.14 μM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Calculate the initial rate, R (often written as V0), of this reaction.

1) Calculate Rmax

The turnover number (Kcat) is a ratio of how many molecules of substrate can be converted into product per catalytic site of a given concentration of enzyme per unit of time:

Kcat = \frac{Vmax}{Et},

where:

Vmax is maximum rate of reaction when all the enzyme sites are saturated with substrate

Et is total enzyme concentration or concentration of total enzyme catalytic sites.

Calculating:

Kcat = \frac{Vmax}{Et}

Vmax = Kcat · Et

Vmax = 4×10^{7} · 1.2 × 10^{-8}

Vmax = 4.8 × 10^{-1} M

2) Calculate the initial rate of this reaction (R):

The Michaelis-Menten equation studies the dynamics of an enzymatic reaction. This model can explain how an enzyme enhances the rate of a reaction and how the reaction rate depends on the concentration of the enzyme and its substrate. The equation is:

V0 = \frac{[S].(Vmax)}{KM + [S]}, where:

[S] is the substrate's concentration

KM is the Michaelis-Menten constant

Substituting [S] = 5.14 × 10^{-6}, KM = 2.5 × 10^{-4} and Vmax = 4.8 × 10^{-1}, the result is V0 = 0.478 M.

The answers are Vmax = 4.8 × 10^{-1} M and V0 = 0.478 M.

 

7 0
3 years ago
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