QuestionCalculate the mass of water added to a beaker that weighed $26.2411 \mathrm{~g} \pm 0.0002 \mathrm{~g}$ and now weighs $33.1932 \mathrm{~g} \pm 0.0002 \mathrm{~g}$ .

Studdy Solution

STEP 1

Assumptions1. The initial weight of the beaker with water is $26.2411 \mathrm{~g} \pm0.000 \mathrm{~g}$ . The final weight of the beaker with added water is $33.193 \mathrm{~g} \pm0.000 \mathrm{~g}$
3. The weight of the beaker itself remains constant and is included in both measurements4. The error associated with each measurement is the given value

STEP 2

First, we need to calculate the weight of the water added. This can be done by subtracting the initial weight of the beaker with water from the final weight of the beaker with added water.
$Weight\, of\, added\, water = Final\, weight - Initial\, weight$

STEP 3

Now, plug in the given values for the initial and final weights to calculate the weight of the added water.
$Weight\, of\, added\, water =33.1932\, g -26.2411\, g$

STEP 4

Calculate the weight of the added water.
$Weight\, of\, added\, water =33.1932\, g -26.2411\, g =6.9521\, g$

STEP 5

Next, we need to calculate the associated error. When subtracting quantities with uncertainties, we add the absolute uncertainties.
$Error = Initial\, error + Final\, error$

STEP 6

Now, plug in the given values for the initial and final errors to calculate the associated error.
$Error =0.0002\, g +0.0002\, g$

STEP 7

Calculate the associated error.
$Error =0.0002\, g +0.0002\, g =0.0004\, g$ So, the weight of the water added to the beaker is $6.9521 \mathrm{~g} \pm0.0004 \mathrm{~g}$ .

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QuestionCalculate the mass of water added to a beaker that weighed $26.2411 \mathrm{~g} \pm 0.0002 \mathrm{~g}$ and now weighs $33.1932 \mathrm{~g} \pm 0.0002 \mathrm{~g}$ .

Studdy Solution

STEP 1

STEP 2

First, we need to calculate the weight of the water added. This can be done by subtracting the initial weight of the beaker with water from the final weight of the beaker with added water.
$Weight\, of\, added\, water = Final\, weight - Initial\, weight$

STEP 3

Now, plug in the given values for the initial and final weights to calculate the weight of the added water.
$Weight\, of\, added\, water =33.1932\, g -26.2411\, g$

STEP 4

Calculate the weight of the added water.
$Weight\, of\, added\, water =33.1932\, g -26.2411\, g =6.9521\, g$

STEP 5

Next, we need to calculate the associated error. When subtracting quantities with uncertainties, we add the absolute uncertainties.
$Error = Initial\, error + Final\, error$

STEP 6

Now, plug in the given values for the initial and final errors to calculate the associated error.
$Error =0.0002\, g +0.0002\, g$

STEP 7

Calculate the associated error.
$Error =0.0002\, g +0.0002\, g =0.0004\, g$ So, the weight of the water added to the beaker is $6.9521 \mathrm{~g} \pm0.0004 \mathrm{~g}$ .

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QuestionCalculate the mass of water added to a beaker that weighed 26.2411 g±0.0002 g26.2411 \mathrm{~g} \pm 0.0002 \mathrm{~g}26.2411 g±0.0002 g and now weighs 33.1932 g±0.0002 g33.1932 \mathrm{~g} \pm 0.0002 \mathrm{~g}33.1932 g±0.0002 g.

Studdy Solution

STEP 1

STEP 2

STEP 3

Now, plug in the given values for the initial and final weights to calculate the weight of the added water.Weight of added water=33.1932 g−26.2411 gWeight\, of\, added\, water =33.1932\, g -26.2411\, gWeightofaddedwater=33.1932g−26.2411g

STEP 4

Calculate the weight of the added water.Weight of added water=33.1932 g−26.2411 g=6.9521 gWeight\, of\, added\, water =33.1932\, g -26.2411\, g =6.9521\, gWeightofaddedwater=33.1932g−26.2411g=6.9521g

STEP 5

Next, we need to calculate the associated error. When subtracting quantities with uncertainties, we add the absolute uncertainties.Error=Initial error+Final errorError = Initial\, error + Final\, errorError=Initialerror+Finalerror

STEP 6

Now, plug in the given values for the initial and final errors to calculate the associated error.Error=0.0002 g+0.0002 gError =0.0002\, g +0.0002\, gError=0.0002g+0.0002g

STEP 7

Calculate the associated error.Error=0.0002 g+0.0002 g=0.0004 gError =0.0002\, g +0.0002\, g =0.0004\, gError=0.0002g+0.0002g=0.0004gSo, the weight of the water added to the beaker is 6.9521 g±0.0004 g6.9521 \mathrm{~g} \pm0.0004 \mathrm{~g}6.9521 g±0.0004 g.

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QuestionCalculate the mass of water added to a beaker that weighed $26.2411 \mathrm{~g} \pm 0.0002 \mathrm{~g}$ and now weighs $33.1932 \mathrm{~g} \pm 0.0002 \mathrm{~g}$ .

Now, plug in the given values for the initial and final weights to calculate the weight of the added water.
$Weight\, of\, added\, water =33.1932\, g -26.2411\, g$

Calculate the weight of the added water.
$Weight\, of\, added\, water =33.1932\, g -26.2411\, g =6.9521\, g$

Next, we need to calculate the associated error. When subtracting quantities with uncertainties, we add the absolute uncertainties.
$Error = Initial\, error + Final\, error$

Now, plug in the given values for the initial and final errors to calculate the associated error.
$Error =0.0002\, g +0.0002\, g$

Calculate the associated error.
$Error =0.0002\, g +0.0002\, g =0.0004\, g$ So, the weight of the water added to the beaker is $6.9521 \mathrm{~g} \pm0.0004 \mathrm{~g}$ .