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Diano4ka-milaya [45]
2 years ago
9

Water is dripping into a bucket at a rate of 51 drops per minute. if each drop has a volume of 0.050 milliliters, what volume of

water will drip into the bucket in 44 minutes?
Chemistry
1 answer:
serious [3.7K]2 years ago
8 0

112.2 milliliters volume of water will drip into the bucket in 44 minutes.

<h3>What is volume?</h3>

How much space an object or substance takes up. • Measured in cubic meters (m3), liters (L) & milliliters (mL).

Total drop = Drops per minute X time

= 51 drops per minute X 44 minutes

=2244 drop

Volume of water will drip into the bucket in 44 minutes

=Total drop X Volume of each drop

=0.050 milliliters X 2244 drop

=112.2 milliliters

Hence, 112.2 milliliters volume of water will drip into the bucket in 44 minutes.

Learn more about volume here:

brainly.com/question/10904074

#SPJ1

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Mass of 2×10^21 number of atoms of an element is 0.4g. What is the mass of 0.5 mole of the element?
alexandr402 [8]

Answer:

66.7 g

Explanation:

Number of atoms = 2×10²¹

Mass of 2×10²¹ atoms = 0.4 g

Mass of 0.5 moles of that element = ?

Solution:

1 mole contain 6.022×10²³ atoms

2×10²¹ atoms × 1 mol / 6.022×10²³ atoms

0.33×10⁻² mol

0.003 mol

 0.003 mole have mass of 0.4 g

0.5 mol have mass 0.5/0.003×0.4 g = 66.7 g

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A 700.0 mL gas sample at STP is compressed to a volume of 200.0 mL, and the temperature is increased to 30.0°C. What is the new
dolphi86 [110]

Answer: The new pressure of the gas in Pa is 388462

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas at STP = 10^5Pa

P_2 = final pressure of gas = ?

V_1 = initial volume of gas = 700.0 ml

V_2 = final volume of gas = 200.0 ml

T_1 = initial temperature of gas = 273 K

T_2 = final temperature of gas = 30^oC=273+30=303K

Now put all the given values in the above equation, we get:

\frac{10^5\times 700.0ml}{273K}=\frac{P_2\times 200.0ml}{303K}

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