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3 years ago

True or Flase: dissolving salt in distilled water creates a homogeneous mixture.

Chemistry

2 answers:

chubhunter [2.5K]3 years ago

7 0

Answer:

true

Explanation:

ser-zykov [4K]3 years ago

3 0

Answer:

true

Explanation:

because the dissolved salt is evenly distributed throughout

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adell [148]

4) would be your correct answer

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2 years ago

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Makovka662 [10]

Answer:

Explanation:

you would have to look more around the page, for example look at some ways that you can right down.

6 0

2 years ago

At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

Alekssandra [29.7K]

Explanation:

$Density =\frac{Mass}[Volume}$

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball = $V=560 cm^3=560 ml=0.560 L$

$D =\frac{0.12 g}{0.560 L}=0.214 g/L$

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = $d_1=1.830 g/L$

Mass of the carbon dioxide gas :

$1.830 g/L\times 0.560 L=1.0248 g$

Total density of filled ball with carbon dioxide gas:

$\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L$

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = $d_2=0.0899 g/L$

Mass of the hydrogen gas :

$1.830 g/L\times 0.560 L=0.050344 g$

Total density of filled ball with hydrogen gas:

$\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L$

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = $d_3=1.330 g/L$

Mass of the oxygen gas :

$1.330 g/L\times 0.560 L=1.7448 g$

Total density of filled ball with oxygen gas:

$\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L$

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = $d_4=1.165 g/L$

Mass of the nitrogen gas :

$1.165 g/L\times 0.560 L=0.6524 g$

Total density of filled ball with nitrogen gas:

$\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L$

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

$\frac{0.12g +m}{0.560L}>1.189 g/L$

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

5 0

2 years ago

What volume in liters of carbon monoxide will be required to produce 18.9 L of nitrogen in the reaction below

mina [271]

Answer:

37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.

Explanation:

Equation for the reaction:

2 CO + 2 NO ------> N2 + 2 CO2

2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen

At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.

So therefore, we can say:

2 * 22.4 L of CO produces 22.4 L of N2

44.8 L of CO produces 22.4 L of N2

Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:

44.8 L of CO = 22.4 L of N

x L = 18.9 L

x L = 18.9 * 44.8 / 22.4

x L = 18.9 * 2

x = 37.8 L

The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L

8 0

3 years ago

Please help me, I really don't want to fail but I don't know how to do this

Brrunno [24]

Answer:

<u>12 moles</u>

Explanation:

$NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)$

__↑______↑

8.00 mol | 14.00 mol

________________

$NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)$

You can turn this into a system of variables which are solvable.

To do this, create variables for the coefficients of each compound in the reaction respectively.

$a(NH_{3}(g)) + b(O_{2}(g)) → \\c(NO_{2}) + d(H_{2}O(g))$

Because to be balanced, the count of atoms in each element of the compound correspond to the coefficient of the variable in that compound so that the count of the left (reactant) side is set equal to the right (product) side.

a corresponds to the coefficient of the first compound, b corresponds to the coefficient of the second compound, c corresponds to the coefficient of the third compound, and d corresponds to the coefficient of the fourth compound.

(Reactant = Product)

Reactant: 1a [N] Product: 1c.

Reactant: 3a [H] Product: 2d.

Reactant: 2b [O] Product: 2c + 1d.

Thus the system is:

1a = 1c

3a = 2d

2b = 2c + 1d.

Then just use the substitution methods to solve.

3 0

2 years ago