True or False Hypoxia results from higher oxygen levels.

Barium + Aluminum nitrate —> makes what ?

Murljashka [212]

Answer:

Aluminum sulfate with barium nitrate to produce burin and aluminum nitrate

5 0

3 years ago

Please someone help it’s due today(I will give brainliest)

k0ka [10]

Answer:

physical or chemical

boiling point

brittleness

opacity

color

density

ductility

shape

hardness

lustre

malleability

melting point

smell

solubility

state of matter

taste

texture

viscosity

Explanation:

3 0

3 years ago

If anyone understands or has this worksheet please help ?!!!!!!!

Drupady [299]

Answer:

Part 1

1. empirical formula is = N₂O₃

2. empirical formula is = NaClO₄

3. empirical formula is = BaCr₂O₇

Part2

no. of atoms of P₄ = 2.1 x 10²³

Part 3

A) no. of moles of S = 0.88 moles

B) no. of atoms of Mg = 1.08 x 10²⁴

C) no. of moles of Br₂ = 9.5 mole

Part 4

A) Molar mass of Na₂SO₄ = 142 g/mol

B) Molar mass of Al₂(SO₄)₃ = 342 g/mol

C) Molar mass of Al₂(SO₄)₃ = 176.5 g/mol

D) Molar mass of K₂CrO₄ = 194 g/mol

Part 5,

mass in grams of I₂ = 254 g

______________

Explanation:

Part 1:

Empirical Formula Calculation from %

1): Data Given

Percent mass of N = 63.6 %

Percent mass of O = 36.4 %

First convert percent to mass

let say we have 100 g of compound

So

mass of N = 63.6 /100 x 100 = 63.6 g

mass of O = 36.4 /100 x 100 = 36.4 g

Now convert masses to moles:

Molar mass of N = 14 g/mol

Molar mass of O = 16 g/mol

Formula used:

no. moles = mass in gram / molar mass ....................(1)

Now find the no. of moles of nitrogen

Put values in formula 1

no. moles = 36.4 g / 14 g/mol

no. moles = 2.6 mol

Now find the no. of moles of Oxygen

Put values in formula 1

no. moles = 63.6 g / 16 g/mol

no. moles = 4 mol

Now calculate the mole ratio of both element

N = 2.6 /2.6 = 1

O = 4 /2.6 = 1.5

To convert the ratio to whole number multiply the ratio with a whole number.

N = 1 x 2 = 2

O = 1.5 x 2 =3

So,

the ratio of N to O 2 : 3 and this is the simplest form

So the empirical formula is = N₂O₃

___________________________________

2): Data Given

Percent mass of Na = 18.8 %

Percent mass of Cl = 29 %

Percent mass of O = 52.3 %

First convert percent to mass

let say we have 100 g of compound

So

mass of Na = 18.8 /100 x 100 = 18.8 g

mass of Cl = 29 /100 x 100 = 29 g

mass of O = 52.3 /100 x 100 = 52.3 g

Now convert masses to moles:

Molar mass of Na = 23 g/mol

Molar mass of Cl = 35.5 g/mol

Molar mass of O = 16 g/mol

Formula used:

no. moles = mass in gram / molar mass ....................(1)

Now find the no. of moles of Na

Put values in formula 1

no. moles = 18.8 g / 23 g/mol

no. moles = 0.82 mol

Now find the no. of moles of Cl

Put values in formula 1

no. moles = 29 g / 35.5 g/mol

no. moles = 0.82 mol

Now find the no. of moles of Oxygen

Put values in formula 1

no. moles = 52.3 g / 16 g/mol

no. moles = 3.3 mol

Calculate the mole ratio of both element

Na = 0.82 / 0.82 = 1

Cl = 0.82 / 0.82 = 1

O = 3.3 / 0.82 = 4

So,

The ratio of Na, Cl and O is 1 : 1 : 4 and this is the simplest form.

So the empirical formula is = NaClO₄

_________________________________

3): Data Given

Percent mass of Ba = 38.9 %

Percent mass of Cr = 29.4 %

Percent mass of O = 31.7 %

First convert percent to mass

let say we have 100 g of compound

So

mass of Ba = 38.9 /100 x 100 = 38.9 g

mass of Cr = 29.4 /100 x 100 = 29 g

mass of O = 31.7 /100 x 100 = 31.7 g

Now convert masses to moles:

Molar mass of Ba = 137 g/mol

Molar mass of Cr = 52 g/mol

Molar mass of O = 16 g/mol

Formula used:

no. moles = mass in gram / molar mass ....................(1)

Now find the no. of moles of Ba

Put values in formula 1

no. moles = 38.9 g / 137 g/mol

no. moles = 0.28 mol

Now find the no. of moles of Cr

Put values in formula 1

no. moles = 29.4 g / 52 g/mol

no. moles = 0.56 mol

Now find the no. of moles of Oxygen

Put values in formula 1

no. moles = 31.7 g / 16 g/mol

no. moles = 2 mol

Calculate the mole ratio of both element

Ba = 0.28 / 0.28 = 1

Cr = 0.56 / 0.28 = 2

O = 2 / 0.28 = 7

So,

The ratio of Ba, Cr and O is 1 : 2 : 7 and this is the simplest form.

So the empirical formula is = BaCr₂O₇

=======================================

****Note: the rest of the answer is in attachment.

5 0

4 years ago

WHAT MASS OF 1,1 DICHLOROEHTANE MUST BE MIXED WITH 100G OF 1,1 DICHLOROTETRAFLUOROEHTANE TO GIVE A SOLUTION WITH VAPOR PRESSURE

vaieri [72.5K]

This is an incomplete question.

The complete question is:

1,1-dichlorotetrafluoroethane, CF3CCL2F, has a vapor pressure of 228 torr. What mass of 1,1-dichloroethane must be mixed with 100.0 g of 1,1-dichlorotetrafluoroethane to give a solution with vapor pressure 157 torr at 25 degrees celsius?

Answer: 46.9 g of 1,1 dichloroethane must be fixed with 100 g of 1,1 dichlorotetrafluoroethane to give a solution with vapor pressure of 157 torr at $25^0C$

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$ = relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

$x_2$ = mole fraction of solute

= $\frac{\text {moles of solute}}{\text {total moles}}$

Given : x g of solute is present in 100 g of solvent

moles of solute (1,1 DICHLOROEHTANE) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{xg}{98.96g/mol}moles$

moles of solvent (1,1 DICHLOROTETRAFLUOROEHTANE ) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{100g}{170.92g/mol}=0.58moles$

Total moles = moles of solute + moles of solvent = $\frac{xg}{98.96g/mol}+0.58$

$x_2$ = mole fraction of solute = $\frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}$

$\frac{228-157}{157}=1\times \frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}$

$0.45=1\times \frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}$

$x=46.9g$

Thus 46.9 g of 1,1 dichloroethane must be fixed with 100 g of 1,1 dichlorotetrafluoroethane to give a solution with vapor pressure of 157 torr at $25^0C$

5 0

4 years ago

1. Consider the unbalanced equation:

Andrei [34K]

Answer:

1.) 17.76 moles H₂O

2.) 323.1 g HCN

Explanation:

1.) An equation is balanced when there is an equal amount of each element on both sides of the reaction. The quantities can be modified by adding coefficients in front of the molecules.

The unbalanced equation:

Al₄C₃(s) + H₂O -----> Al(OH)₃(s) + CH₄(g)

Reactants: 4 aluminum, 3 carbon, 2 hydrogen, 1 oxygen

Products: 1 aluminum, 1 carbon, 7 hydrogen, 3 oxygen

The balanced equation:

Al₄C₃(s) + 12 H₂O -----> 4 Al(OH)₃(s) + 3 CH₄(g)

Reactants: 4 aluminum, 3 carbon, 24 hydrogen, 12 oxygen

Products: 4 aluminum, 3 carbon, 24 hydrogen, 12 oxygen

Now that the equation is balanced, you can use the relevant coefficients to construct a mole-to-mole ratio. This will allow you to convert between moles Al₄C₃ to moles H₂O.

1.48 moles Al₄C₃ 12 moles H₂O
---------------------------- x ------------------------ = 17.76 moles H₂O
1 mole Al₄C₃

2.) To find the mass of HCN, you need to (1) convert grams NH₃ to moles NH₃ (via molar mass), then (2) convert moles NH₃ to moles HCN (via mole-to-mole from equation coefficients), and then (3) convert moles HCN to grams HCN (via molar mass). It is important to arrange the coefficients in a way that allows for the cancellation of units.

Molar Mass (NH₃): 14.009 g/mol + 3(1.008 g/mol)

Molar Mass (NH₃): 17.033 g/mol

Molar Mass (HCN): 1.008 g/mol + 12.011 g/mol + 14.009 g/mol

Molar Mass (HCN): 27.028 g/mol

2 CH₄(g) + 3 O₂(g) + 2 NH₃(g) ------> 2 HCN(g) + 6 H₂O(g)

203.6 g NH₃ 1 mole 2 moles HCN 27.028 g
--------------------- x ---------------- x ------------------------ x ----------------- =
17.033 g 2 moles NH₃ 1 mole

= 323.1 g HCN

8 0

2 years ago