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alexgriva [62]
3 years ago
14

Conversación de masa 3,6kg a g​

Chemistry
1 answer:
Nikolay [14]3 years ago
7 0

Explanation:

1kg = 1000g

3.6 kg = 3.6 × 100

3600g

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A sample of aluminum is placed in a 25 ml graduated cylinder containing 10 mL of water. The level of water rises to 18 mL. Alumi
anastassius [24]

To determine the mass of the sample, first find the volume difference after and before the aluminum was placed, the volume change is equal to the volume of the submerged object, in this case aluminum.

Then knowing volume of aluminum and the density of it, we can solve for the mass.

D = m/v

Dv = m

2.7 g/ml • 8 ml = 21.6 grams.

4 0
3 years ago
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The following unbalanced equation illustrates the overall reaction by which the body utilizes glucose to produce energy: C6H12O6
s344n2d4d5 [400]

Answer:

the conversion factor is f= 6  mol of glucose/ mol of CO2

Explanation:

First we need to balance the equation:

C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)

the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:

f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction

f = 6 moles of CO2 / 1 mol of glucose = 6  mol of glucose/ mol of CO2

f = 6 mol of CO2/ mol of glucose

for example, for 2 moles of glucose the number of moles of CO2 produced are

n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2

3 0
3 years ago
C5H12 + 8O2 → 5CO2 + 6H2O<br><br> Calculate the energy change for the reaction:
KATRIN_1 [288]
C5H12 (l) + 8O2 (g) ----> 5CO2 (g) + 6H2O (l)
Delta H = -3505.8 kJ/mol

C (s) + O2 (g) -----> CO2 (g)
Delta H = -393.5 kJ/mol

H2 (g) + (1/2)O2 (g) ------> H2O (l)
Delta H = -286 kJ/mol

Possible answers:
a. +35 kJ/mol
b. + 1,073 kJ/mol
c. -4,185 kJ/mol
d. -2,826 kJ/mol
e. -178 kJ/mol
8 0
3 years ago
A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo
Liula [17]

<u>Answer:</u> The molar mass of the insulin is 6087.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g   (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = 62.364\text{ L.mmHg }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol

Hence, the molar mass of the insulin is 6087.2 g/mol

8 0
3 years ago
When liquid water evaporates to gaseous water
alina1380 [7]

Answer: liquid water 2

Explanation:

5 0
3 years ago
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