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Darina [25.2K]
2 years ago
15

A student measures the volume of a solution to be 0.01370 l. How many significant digits are in this measurement?

Chemistry
1 answer:
Mkey [24]2 years ago
7 0

A student measures the volume of a solution to be 0.01370 have 5 significant digits in this measurement.

<h3>What are significant digits?</h3>

The significant digits are the minimum number from zero to nine for reporting any measurement where the digits are uncertain.

The significant digits starting from zero are not significant digits, decimal is not a significant digit, and ending zero after the decimal are significant digits.

Therefore, the student measures the volume of a solution to be 0.01370 5 significant digits are in this measurement.

Learn more about significant digits, here:

brainly.com/question/1658998

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A gas occupies a volume at 34.2 mL at a temperature of 15.0 C and a pressure of 800.0 torr. What will be the volume of this gas
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The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
     P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
     (800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
     V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
     V2 = 34.1 mL
3 0
3 years ago
Read 2 more answers
At 20°C the vapor pressure of benzene (C6H6) is 75 torr, and that of toluene (C7H8) is 22 torr. Assume that benzene and toluene
valina [46]

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Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

p_1=x_1p_1^0 and p_2=x_2P_2^0

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According to Dalton's law, the total pressure is the sum of individual pressures.

p_{total}=p_1+p_2\\p_{total}=x_{benzene}p_{benzene}^0+x_{toluene}P_{toluene}^0

x_{benzene}=x,

x_{toluene}=1-x_{benzene}=1-x,

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Thus (1-x0 = (1-0.34)=0.66

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