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1 The mass of a gold ring is 30 grams (gr), and the volume is 1.5 cm 3; WhichIs it the density of the ring?2 A bag of compost has a mass of 2 Kg. Express the mass in grams3 Express the mass of an orange weighing 138 g in units of kg4 The content of a soda bottle is 250 milliliters (ml). Express thevolume in liters.Measurement units:1 kilograms (kg) = 1000 grams (gr)1 liter (i) = 1000 cm 3 or ml |
| Answer» <html><body><p><strong>Answer:</strong></p><p>1 d = </p><p>m</p><p>V</p><p> </p><p></p><p>m = d×V</p><p></p><p>V = </p><p>m</p><p>d</p><p> </p><p></p><p>DENSITY</p><p></p><p>Density is defined as mass per unit volume.</p><p></p><p>d = </p><p>m</p><p>V</p><p> </p><p></p><p>Example:</p><p>A brick of salt measuring 10.0 cm x 10.0 cm x 2.00 cm has a mass of 433 g. What is its density?</p><p></p><p>Step 1: Calculate the volume</p><p>V = lwh = 10.0 cm × 10.0 cm × 2.00 cm = 200 cm³</p><p></p><p>Step 2: Calculate the density</p><p></p><p>d = </p><p>m</p><p>V</p><p> = </p><p>433</p><p>g</p><p>200</p><p>c</p><p>m</p><p>³</p><p> = 2.16 g/cm³</p><p></p><p>MASS</p><p></p><p>d = </p><p>m</p><p>V</p><p> </p><p></p><p>We can rearrange this to get the expression for the mass.</p><p></p><p>m = d×V</p><p></p><p>Example:</p><p>If 500 mL of a liquid has a density of 1.11 g/mL, what is its mass?</p><p></p><p>m = d×V = 500 mL × </p><p>1.11</p><p>g</p><p>1</p><p>m</p><p>L</p><p> = 555 g</p><p></p><p>VOLUME</p><p></p><p>d = </p><p>m</p><p>V</p><p> </p><p></p><p>We can rearrange this to get the expression for the volume.</p><p></p><p>V = </p><p>m</p><p>d</p><p> </p><p></p><p>Example:</p><p>What is the volume of a bar of gold that has a mass of 14.83 kg. The density of gold is 19.32 g/cm³.</p><p></p><p>Step 1: Convert kilograms to grams.</p><p></p><p>14.83 kg × </p><p>1000</p><p>g</p><p>1</p><p>k</p><p>g</p><p> = 14 830 g</p><p></p><p>Step 2: Calculate the volume.</p><p></p><p>V = </p><p>m</p><p>d</p><p> = 14 830 g × </p><p>1</p><p>c</p><p>m</p><p>³</p><p>19.32</p><p>g</p><p> = 767.6 cm³</p><p></p><p> Ernest Z. · 32 · Jan 11 2014</p><p>How can density be used to <a href="https://interviewquestions.tuteehub.com/tag/identify-496703" style="font-weight:bold;" target="_blank" title="Click to know more about IDENTIFY">IDENTIFY</a> substances?</p><p>You can identify an unknown substance by measuring its density and comparing your result to a list of known densities.</p><p></p><p>Density = mass/volume. Assume that you have to identify an unknown metal. You can determine the mass of the metal on a scale. You can determine the volume by dropping the <a href="https://interviewquestions.tuteehub.com/tag/object-11416" style="font-weight:bold;" target="_blank" title="Click to know more about OBJECT">OBJECT</a> into a graduated cylinder containing a known volume of water and measuring the new volume. You divide the mass by the volume and compare the density to a list of known densities.</p><p></p><p>EXAMPLE</p><p></p><p>A metal bolt with a mass of 99.7 g is <a href="https://interviewquestions.tuteehub.com/tag/dropped-2594665" style="font-weight:bold;" target="_blank" title="Click to know more about DROPPED">DROPPED</a> into a graduated cylinder containing 50.0 cm³ of water. The new volume reads</p><p>72.1 cm³. Identify the metal.</p><p></p><p>Solution</p><p></p><p>V = 72.1 cm³ - 50.0 cm³ = 22.1 cm³</p><p></p><p>D = </p><p>m</p><p>V</p><p>=</p><p>99.7</p><p>g</p><p>22.1</p><p>c</p><p>m</p><p>³</p><p> = 4.51 g/cm³</p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p>2HI, If you are using certain <a href="https://interviewquestions.tuteehub.com/tag/amount-374803" style="font-weight:bold;" target="_blank" title="Click to know more about AMOUNT">AMOUNT</a> of given fertilizer for your experiment then you must have added certain dose <a href="https://interviewquestions.tuteehub.com/tag/lets-541222" style="font-weight:bold;" target="_blank" title="Click to know more about LETS">LETS</a> say x% and then you might have use different doses for other treatment like 75% of x, 50% of x and so on. If this is the case you can easily calculate required amount of dose for your given field area</p><p>Just convert the required area into m2 (suppose for one acre you got 4,000m2). At the same time convert the experimental plot or pot into m2. Then figure out the most responsive fertilizer % which you want to use in real field. Lets say if area is 1m2 for the experiment unit and you get good response when you used 200gm of fertilizer ( suppose having concentration of (10%, 10%,10% NPK).</p><p>You can get your required amount 200 gm *4,000m2.</p><p>Hope this will help. But I am not entirely sure how did you use the fertilizer for your response experiment</p><p></p><p></p><p></p><p></p><p></p><p></p><p>sorry i can answer this much only</p></body></html> | |