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True [87]
3 years ago
14

1. You may "feel" cold when you touch certain kinds of matter because​

Chemistry
1 answer:
Vinil7 [7]3 years ago
8 0
You may feel cold when you touch certain kinds of matter because some matter gives off cold air. I’m NOT SURE
You might be interested in
Trying to take the ged test please help me I’m stuck
JulsSmile [24]

Answer: C) Either benzene or oxygen may limit the amount of product that can be formed

Explanation: Benzene and oxygen are the reactants of the equation. What type and the amount of reactants there are in a chemical reaction affects the outcome. Therefore, seeing as benzene and oxygen are the reactants, the answer is C).

3 0
2 years ago
How many water molecules does it contain? The density of water is 1.0 g/cm3.
max2010maxim [7]

Complete question is;

A drop of water has a volume of approximately 7 × 10⁻² ml. How many water molecules does it contain? The density of water is 1.0 g/cm³.

This question will require us to first find the number of moles and then use avogadro's number to get the number of water molecules.

<em><u>Number of water molecules = 2.34 × 10²¹ molecules</u></em>

We are given;

Volume of water; V = 7 × 10⁻² ml

Density of water; ρ = 1 g/cm³ = 1 g/ml

Formula for mass is; m = ρV

m = 1 × 7 × 10⁻²

m = 7 × 10⁻² g

from online calculation, molar mass of water = 18.01 g/mol

Number of moles(n) = mass/molar mass

Thus;

n = (7 × 10⁻²)/18.01

n = 3.887 × 10⁻³ mol

from avogadro's number, we know that;

1 mol = 6.022 × 10²³ molecules

Thus,3.887 × 10⁻³ mol will give; 6.022 × 10²³ × 3.887 × 10⁻³ = 2.34 × 10²¹ molecules

Read more at; brainly.in/question/17990661

6 0
2 years ago
(4) Calculate the % of a compound that can be removed from liquid phase 1 by using ONE to FOUR extractions with a liquid phase 2
maksim [4K]

Answer:

One extraction: 50%

Two extractions: 75%

Three extractions: 87.5%

Four extractions: 93.75%

Explanation:

The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.

qⁿ = (V₁/(V₁ + KV₂))ⁿ

We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:

qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ =  (V₁/(2V₁))ⁿ = (1/2)ⁿ

When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.

When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.

When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.

When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.

5 0
3 years ago
A 25.0 ml sample of 0.150 m hydrazoic acid is titrated with a 0.150 m naoh solution. what is the ph after 15.0 ml of the sodium
babunello [35]
First, we need to calculate moles of hydrazoic acid NH3:

moles NH3 = molarity * volume 

                    = 0.15 m * 0.025 L

                   =  0.00375 moles

moles NaOH = molarity * volume 

                       = 0.15 m * 0.015 L

                       = 0.00225 moles 

after that we shoul get the total volume = 0.025L + 0.015L

                                                                   = 0.04 L

So we can get the concentration of NH3 & NaOH by:

∴[NH3] = moles NH3 / total volume 

           = 0.00375 moles / 0.04 L

           = 0.09375 M

∴[NaOH] = moles NaOH / total volume 

                = 0.00225 moles / 0.04 L

                = 0.05625 M

then, when we have the value of Ka of NH3 so we can get the Pka value from:

Pka = -㏒Ka 

       = - ㏒ 1.9 x10^-5

      = 4.7 

finally, by using H-H equation we can get PH:

PH = Pka + ㏒[salt/ basic]

PH = 4.7 +㏒[0.05625/0.09375]

∴ PH = 4.48 


6 0
2 years ago
The following initial rate data are for the reaction of hypochlorite ion with iodide ion in 1M aqueous hydroxide solution: OCI+r
Vinil7 [7]

Answer:

Rate = k [OCl] [I]

Explanation:

OCI+r → or +CI

Experiment [OCI] M I(-M) Rate (M/s)2

1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3

2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3

3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3

4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3

The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.

In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.

In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.

The rate law is given as;

Rate = k [OCl] [I]

5 0
2 years ago
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