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3 years ago

If a drop of blood is 0.05 mL, how many drops of blood are in a blood collection tube that holds 2 mL ?

Chemistry

2 answers:

soldi70 [24.7K]3 years ago

7 0

Just divide the two (2 / 0.05) and you will get your answer; there are 40 drops of bloodin the collection tube.

Drupady [299]3 years ago

7 0

Answer : The number of drops of blood in a blood collect in tube are, 40.

Explanation :

As we are given that:

Volume of a drop of blood = 0.05 mL

Volume of blood collection tube that holds = 2 mL

Now we have to calculate the number of drops of blood collect in tube.

$\text{Number of drops of blood}=\frac{\text{Volume of blood collect in a tube}}{\text{Volume of a drop of blood}}$

Now put all the given values in this expression, we get:

$\text{Number of drops of blood}=\frac{2mL}{0.05mL}=40$

Therefore, the number of drops of blood in a blood collect in tube are, 40.

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C

ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)

The enthalpy of reaction (1) is given by:

$\Delta H = \Delta H_{r} - \Delta H_{p}$ (2)