question 1\nthe temperature of 25°c is ____ in kelvins.\n103\n298\n166\n248\n138\n\n1 pts\nquestion 2\nthe…

question 1\nthe temperature of 25°c is ____ in kelvins.\n103\n298\n166\n248\n138\n\n1 pts\nquestion 2\nthe density of a gold nugget is 19.3 g/cm³. if the volume of the gold nugget is 0.00369 l, the mass of the nugget is ____ g.\n5.23\n19.3\nnone of the above\n71.2\n0.191

question 1\nthe temperature of 25°c is ____ in kelvins.\n103\n298\n166\n248\n138\n\n1 pts\nquestion 2\nthe density of a gold nugget is 19.3 g/cm³. if the volume of the gold nugget is 0.00369 l, the mass of the nugget is ____ g.\n5.23\n19.3\nnone of the above\n71.2\n0.191

Answer

Explanation:

Question 1

Step1: Recall temperature - conversion formula

The formula to convert Celsius to Kelvin is $K = ^{\circ}C+ 273.15$.

Step2: Substitute the given Celsius value

Given $^{\circ}C = 25$, then $K=25 + 273.15=298.15\approx298$.

Question 2

Step1: Convert volume to $cm^{3}$

Since $1L = 1000cm^{3}$, for a volume of $V = 0.00369L$, then $V=0.00369\times1000 = 3.69cm^{3}$.

Step2: Use the density - mass formula

The density formula is $\rho=\frac{m}{V}$, where $\rho$ is density, $m$ is mass and $V$ is volume. Rearranging for mass gives $m=\rho V$. Given $\rho = 19.3g/cm^{3}$ and $V = 3.69cm^{3}$, then $m=19.3\times3.69 = 71.217\approx71.2g$.

Answer:

Question 1: B. 298 Question 2: D. 71.2