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NikAS [45]
3 years ago
8

Properties of a solution that depend only on the ratio of the number of particles of solute and solvent in the solution are know

n as:
intensive properties

extensive properties

colligative properties

chemical properties
Chemistry
1 answer:
aleksandrvk [35]3 years ago
5 0
Properties of a solution that depend only on the ratio of the number of particles of solute and solvent in the solution are known as <span>colligative properties</span>
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Help please! I'd appreciate it 
babymother [125]

Answer:

16.56 g

Explanation:

Mass is the production of Volume and Density.

m = V. d = 6 × 2.76 = 16.56 g

6 0
3 years ago
How many moles of cf4 are there in 171g of cf4 ?
xeze [42]
1 is + 4  = to n please
5 0
3 years ago
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A blank is a condition that strays from normal homeostasis.
AURORKA [14]

homeostatic imbalance is the answer, because it's when the internal environment cannot remain in equilibrium.

8 0
4 years ago
An empty container weighs 80.21 g. it is filled with 20.14 mL of an unknown liquid, and the total weight of the container and li
tatiyna

Answer:

1.242 g/mL

Explanation:

Step 1: Given data

Mass of the empty container (m₁): 80.21 g

Mass of the filled container (m₂): 105.22 g

Volume of the unknown liquid (V): 20.14 mL

Step 2: Calculate the mass of the liquid

The mass of the liquid is equal to the difference between the mass of the filled container and the mass of the empty container.

m = m_2 - m_1 = 105.22g - 80.21 g = 25.01 g

Step 3: Calculate the density of the unknown liquid

The density of the liquid is equal to its mass divided by its volume.

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4 0
3 years ago
The data below shows the change in concentration of dinitrogen pentoxide over time, at 330 K, according to the following process
tensa zangetsu [6.8K]

Answer: a) 1.7\times 10^{-4}

b) 3.4\times 10^{-4}

Explanation:

The reaction is :

2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)

Rate = Rate of disappearance of N_2O_5 = Rate of appearance of NO_2

Rate =  -\frac{d[N_2O_5]}{2dt} = \frac{d[NO_2]}{4dt}

Rate of disappearance of N_2O_5 = \frac{\text {change in concentration}}{time} = \frac{0.100-0.066}{200-0}=1.7\times 10^{-4}

a) Rate of disappearance of N_2O_5 = -\frac{d[N_2O_5]}{2dt}

Rate of appearance of NO_2 = \frac{d[NO_2]}{4dt}

b) Rate of appearance of NO_2 =  \frac{d[NO_2]}{dt}=2\times 1.7\times 10^{-4}}=3.4\times 10^{-4}

8 0
3 years ago
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