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mihalych1998 [28]
3 years ago
14

What is the slope of a line that is perpendicular to the line shown?

Chemistry
2 answers:
Nonamiya [84]3 years ago
7 0
I think the answer is 3
adell [148]3 years ago
4 0
3

the slope of the line shown is -1/3, so the perpendicular line is the opposite reciprocal which would be 3
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Comment on whether each of the following is a homogeneous mixture or a heterogeneous mixture: (a) air in a closed bottle, (b) ai
Colt1911 [192]

Explanation:

A mixture in which there is uniform distribution of solute particles into the solvent is known as a homogeneous mixture.

For example, sugar dissolved in water is a homogeneous mixture.

On the other hand, a mixture in which there is uneven distribution of solute particles into the solvent is known as a heterogeneous mixture.

For example, sand present in water is a heterogeneous mixture.

Comment on given situations will be as follows.

(a)   Air in a closed bottle - It is a homogeneous mixture because there will be even distribution of other gases that are present in air.

(b)   Air over New York City - It is a heterogeneous mixture because there will be presence of some dust particles, fog or smoke into the air. Distribution of all these particles will be uneven. This will make air over New York City heterogeneous in nature.

8 0
3 years ago
When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t
OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

Mn(s)+2HCl(aq)\rightarrow  MnCl_2(aq)+H_2(g)

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

m=1.00 g/mL\times 100.0 mL = 100 g

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

q=m\times c\times (T_{final}-T_{initial})

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = 4.18 J/^oC

T_{final} = final temperature = 23.1^oC

T_{initial} = initial temperature = 28.9^oC

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

q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC

q=2,242.4 J=2.242 kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = \frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol

\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0
2 years ago
Describe how you would crystallise potassium nitrate from its aqueous liquid?
Rus_ich [418]
Just look here https://www.jiskha.com/search/index.cgi?query=describe+how+you+can+crystallize+potassium+nitrate+fro...
5 0
3 years ago
I’m so lost please help
ExtremeBDS [4]

C=4

D=-19

Explanation:

2exp-25/5exp-7= 4exp-19

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wel

Answer:

DANG :/ IM TOO LATE

Explanation:

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