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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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Any element can always be identified by its:
Alborosie

Explanation:

atomic mass

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3 0
3 years ago
Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. 0.100 mole of Ca(N
pickupchik [31]

Answer:

[Ca²⁺] = 1M

[NO₃⁻] = 2M

Explanation:

Calcium nitrate dissociates in water as follows:

Ca(NO₃)₂ ⇒ Ca²⁺ + 2NO₃⁻

The moles of Ca²⁺ can be found using the molar relationship between Ca(NO₃)₂ and Ca²⁺

(0.100mol Ca(NO₃)₂) (Ca²⁺ /Ca(NO₃)₂) = 0.100 mol Ca²⁺

The concentration of Ca²⁺  is then:

[Ca²⁺] = n/V = (0.100mol)/(100.0mL) x (1000ml)/(1L) = 1M

Similarly, moles of NO₃⁻ can be found using the molar relationship between Ca(NO₃)₂ and NO₃⁻:

(0.100mol Ca(NO₃)₂) (2NO₃⁻/Ca(NO₃)₂) = 0.200 mol NO₃⁻

The concentration of NO₃⁻ is then:

[NO₃⁻] = (0.200mol)/(100.0mL) x (1000ml)/(1L) = 2M

6 0
3 years ago
Is Al2O3 an acid or a base?
kirza4 [7]
 Al_2O_3 is aluminium oxide (also called aloxide, aloxite and alundum). It is neither acid nor base. It has amphoteric nature meaning it can act as an acid with bases and as a base with acids. It neutralises them to salts. 
4 0
3 years ago
Read 2 more answers
What is the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen
natta225 [31]
Carbon(C):
number of moles= mass/molar mass(Mr)
=65.5/12
=5.5 moles

Hydrogen(H):
number of moles=mass/molar mass (Mr)
=5.5/1
=5.5 moles

Oxygen (O):
number of moles = mass/molar mass (Mr)
=29.0/16
=1.8 moles

EF= lowest number of moles over each of the elements

So,
C= 5.5/1.8 = 3
H= 5.5/1.8 = 3
O= 1.8/1.8 = 1

Therefore Emperical formula= C3H3O
6 0
2 years ago
1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.
Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice )  = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}

\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}

\Delta \ S = {(1*37.7)In \dfrac{263}{273}

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe  >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0
3 years ago
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