PLEASE HELP!!! What are 3 types of heat energy transfer?

Are all elements abbreviated with one or two letters? If not, name some that aren't.

mixas84 [53]

Some elements that are not abbreviated with one or two letters on the periodic table are Ununoctium (Uuo) and Ununtrium (Uut).

7 0

4 years ago

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When heated a sample consisting of only CaCO3 and MgCO3 yields a mixture of CaO and Mgo. If the weight of the combined oxides is

ExtremeBDS [4]

Answer:

38.83 % of CaCO3

61.17 % of MgCO3

Explanation:

where Moles of CaCO3 is equals to x and MgCO3 is y we have that...

CaCO3 molar mass = 100.09 g / mol = 100.09 x

MgCO3 molar mass = 84.31 g / mol = 84.31 y

decomposition reactions :

CaCO3 ---> CaO + CO2

MgCO3 ---> MgO + CO2

So we have that , Moles of CaO = Moles of CaCO3 = x

and Moles of MgO = Moles of MgCO3 = y

CaO molar mass = 56.08 g / mol

MgO molar mass = 40.30 g / mol

CaO = 56.08 x

MgO = 40.30 y

"If the weight of the combined oxides is equal to 51.00% of the initial sample weight,"

total mass of MgO and CaO = 51.00 % of Total Mass of MgCO3 and CaCO3

thus

56.08 x + 40.30 y = 0.51 ( 100.09 x + 84.31 y )

56.08 x + 40.30 y = 51.04 x + 42.99y

5.04 x = 2.7 y

y = 1.87 x

CaCO3 % in the sample

= 100.09 x × 100 / ( 100.09 x + 84.31 y )

= 10009 x / ( 100.09 x + 84.31 × 1.87 x )

= 10009 x / ( x ( 100.09 + 157.66 ) )

= 10009 / 257.75

= 38.83 %

MgCO3 % in the sample

= 100 - 38.83

= 61.17 %

7 0

3 years ago

Commercial hydrochloric acid (HCl) is typically labeled as being 38.0 % (weight %). The density of HCl is 1.19 g/mL. a) What is

Strike441 [17]

Answer:

For a: The molarity of commercial HCl solution is 12.39 M.

For b: The molality of commercial HCl solution is 16.79 m.

For c: The volume of commercial HCl solution needed is 2.42 L.

Explanation:

We are given:

Mass % of commercial HCl solution = 38 %

This means that 38 grams of HCl is present in 100 grams of solution.

To calculate the volume of solution, we use the equation:

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

Density of HCl solution = 1.19 g/mL

Mass of solution = 100 g

Putting values in above equation:

$1.19g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=84.034mL$

For a:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = ?

Molar mass of HCl = 36.5 g/mol

Volume of solution = 84.034 mL

Mass of HCl = 38 g

Putting values in above equation, we get:

$\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 84.034}\\\\\text{Molality of commercial HCl solution}=12.39M$

Hence, the molarity of commercial HCl solution is 12.39 M.

For b:

To calculate the molality of solution, we use the equation:

$Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (HCl) = 38 g

$M_{solute}$ = Molar mass of solute (HCl) = 36.5 g/mol

$W_{solvent}$ = Mass of solvent = 100 - 38 = 62 g

Putting values in above equation, we get:

$\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 62}\\\\\text{Molality of commercial HCl solution}=16.79m$

Hence, the molality of commercial HCl solution is 16.79 m.

For c:

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution

We are given:

$M_1=6M\\V_1=5.00L\\M_2=12.39M\\V_2=?L$

Putting values in above equation, we get:

$6\times 5=12.39\times V_2\\\\V_2=2.42L$

Hence, the volume of commercial HCl solution needed is 2.42 L.

5 0

3 years ago

Please help me asap!!

NNADVOKAT [17]

Answer:

Al2o3 is 101

(nh4)2O IS 52

S8 is 256.56

Ba(oh)2 is 171.35

Cacl2 is 110.98

H2O is 18.01

Explanation:

5 0

3 years ago

How many significant digits are in 12,003,000 A) 8 B) 5 C) 6 D) 3

Ksju [112]

Answer:5

Explann:

I've actually not learnt it but i can solve other questions under Maths

5 0

4 years ago