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3 years ago

How many megagrams are there in 7.020 * 10^-10 nanograms

2 answers:

7 0

<span>7.020 * 10^-25
There are 15 orders of magnitude between megagrams and nanograms Megagram = 1 x 10^6 gram
Nanogram = 1 x 10^-9 gram</span>

Ilia_Sergeevich [38]3 years ago

5 0

From the basics conversions:
1 nanogram is equivalent to 1 * 10^-15 megagrams

To convert 7.02 * 10^-10 nanograms to megagrams, all you have to do is cross multiplication as follows:
1 nanogram ................> 1 * 10^-15 megagrams
7.02 * 10^-10 nanograms ...............> ?? megagrams

megagrams = (7.02*10^-10 * 1*10^-15) / (1) = 7.02 * 10^-25 megagrams

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3 years ago

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

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When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

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$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

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2 years ago

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3 years ago

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2 years ago