Question

question 6
the number of cubic centimeters (cm³) in 43.0 ml is
a 4.30 cm³
b 43.0 cm³
c 0.0430 cm³
d none of the above
question 7
how many grams are in 1.10 - lb sample of iron(ii) phosphate? (1 lb = 453.6 g)
a 499 g
b 1.10 g
c 2.43 g
d 0.499 g
e 2.00 g
question 8

Explanation:

Step1: Unit conversion relationship

We know that \(1\space mL = 1\space cm^{3}\).

Step2: Substitute the value

Given the volume \(V = 43.0\space mL\), using the conversion \(V(cm^{3})=V(mL)\times1\), we substitute \(V = 43.0\space mL\) into the formula. So \(V=43.0\times1\space cm^{3}=43.0\space cm^{3}\)

Step1: Use the given conversion factor

We are given that \(1\space lb = 453.6\space g\). To convert the mass of the iron(II) phosphate sample from pounds to grams, we use the formula \(m(g)=m(lb)\times453.6\space g/lb\)

Step2: Substitute the value

Substitute \(m = 1.10\space lb\) into the formula: \(m=1.10\times453.6\space g\).

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Answer:

B. \(43.0\space cm^{3}\)