QUESTION IMAGE
Question
all densities are at 298 k (25°c) unless otherwise noted.
you can enter your answers in decimal notation, or you can enter them in scientific notation using e or x 10^ for the
power of 10 (e.g., enter 6.02 x 10^{23} as 6.02e23 or 6.02x10^23).
- given: d_{indium}=7.3 g/cm^{3}, 0.0470 cm^{3}
wanted: mass of indium in grams?
g
- given: d_{phosphorus}=1.82 g/cm^{3}, 2.19 g
wanted: volume of phosphorus in cm^{3}?
cm^{3}
- given: europium, 4.62 g, 0.888 cm^{3}
wanted: density of europium in g/cm^{3}?
g/cm^{3}
- given: d_{osmium}=22.6 g/cm^{3}, 0.149 cm^{3}
wanted: mass of osmium in grams?
g
- given: d_{glycerol}=1.261 g/ml, 0.408 ml
wanted: mass of glycerol in grams?
g
- given: rhodium, 0.393 g, 0.03169 cm^{3}
wanted: density of rhodium in g/cm^{3}?
g/cm^{3}
- given: d_{xenon}=0.0059 g/ml at stp, 510 ml
wanted: mass of xenon in grams?
g
- given: d_{aerogel}=0.001000 g/cm^{3}, 2100 cm^{3}
wanted: mass of aerogel in grams?
g
Step1: Use the formula \(m = D\times V\) (mass = density × volume)
For problem 1: \(m_{indium}=D_{indium}\times V_{indium}=7.3\times0.0470 = 0.3431\) g
For problem 4: \(m_{osmium}=D_{osmium}\times V_{osmium}=22.6\times0.149 = 3.3674\) g
For problem 5: \(m_{glycerol}=D_{glycerol}\times V_{glycerol}=1.261\times0.408 = 0.514488\) g
For problem 7: \(m_{xenon}=D_{xenon}\times V_{xenon}=0.0059\times510 = 3.009\) g
For problem 8: \(m_{aerogel}=D_{aerogel}\times V_{aerogel}=0.001000\times2100 = 2.1\) g
Step2: Use the formula \(V=\frac{m}{D}\) (volume = mass / density)
For problem 2: \(V_{phosphorus}=\frac{m_{phosphorus}}{D_{phosphorus}}=\frac{2.19}{1.82}\approx1.2033\) \(cm^{3}\)
Step3: Use the formula \(D = \frac{m}{V}\) (density = mass / volume)
For problem 3: \(D_{europium}=\frac{m_{europium}}{V_{europium}}=\frac{4.62}{0.888}\approx5.2027\) \(g/cm^{3}\)
For problem 6: \(D_{rhodium}=\frac{m_{rhodium}}{V_{rhodium}}=\frac{0.393}{0.03169}\approx12.4\) \(g/cm^{3}\)
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- \(0.3431\) g
- \(1.2033\) \(cm^{3}\)
- \(5.2027\) \(g/cm^{3}\)
- \(3.3674\) g
- \(0.514488\) g
- \(12.4\) \(g/cm^{3}\)
- \(3.009\) g
- \(2.1\) g