Question

question 13
the molar mass of c₂h₄(nh₂)₃ is blank 1 g/mol.
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question 14
if a box of cacl₂ has a mass of 2.56 g. blank 1 moles are in the box. blank 2x10^blank 3 atoms are in the box.
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blank 2 add your answer
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question 15
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Explanation:

Step1: Calculate molar mass of \(C_2H_4(NH_2)_3\)

The molar - mass of carbon (\(C\)) is approximately \(12.01\ g/mol\), hydrogen (\(H\)) is approximately \(1.01\ g/mol\), and nitrogen (\(N\)) is approximately \(14.01\ g/mol\).
In \(C_2H_4(NH_2)_3\), there are \(2\) carbon atoms, \((4 + 2\times3)=10\) hydrogen atoms, and \(3\) nitrogen atoms.
The molar mass \(M\) is \(M = 2\times12.01+10\times1.01 + 3\times14.01\)
\[

$$\begin{align*} M&=24.02+10.1+42.03\\ &=76.15\ g/mol \end{align*}$$

\]

Step2: Calculate moles of \(CaCl_2\)

The molar mass of \(CaCl_2\): The molar - mass of calcium (\(Ca\)) is approximately \(40.08\ g/mol\), and the molar - mass of chlorine (\(Cl\)) is approximately \(35.45\ g/mol\). The molar mass of \(CaCl_2\) is \(M_{CaCl_2}=40.08 + 2\times35.45=40.08+70.9 = 110.98\ g/mol\).
The number of moles \(n\) of \(CaCl_2\) with a mass \(m = 2.56\ g\) is \(n=\frac{m}{M_{CaCl_2}}=\frac{2.56}{110.98}\approx0.0231\) moles.

Step3: Calculate number of atoms in \(CaCl_2\)

In one formula - unit of \(CaCl_2\), there are \(3\) atoms (\(1\) \(Ca\) and \(2\) \(Cl\)).
The number of formula - units of \(CaCl_2\) is \(N_{f}=n\times N_A\), where \(N_A = 6.022\times10^{23}\ mol^{-1}\).
\(N_{f}=0.0231\times6.022\times10^{23}\approx1.391\times10^{22}\) formula - units.
The number of atoms \(N\) is \(N = 3\times N_{f}=3\times1.391\times10^{22}=4.173\times10^{22}\) atoms, which can be written as \(4.17\times10^{22}\) (rounded to two decimal places).

Answer:

Blank 1 (for \(C_2H_4(NH_2)_3\) molar mass): \(76.15\)
Blank 1 (for \(CaCl_2\) moles): \(0.0231\)
Blank 2 (for \(CaCl_2\) atoms coefficient): \(4.17\)
Blank 3 (for \(CaCl_2\) atoms exponent): \(22\)