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3 years ago

What is the molar mass of a compound if 25.0 g of the compound dissolved in 750. Ml gives a molarity of 0.290 m?

Chemistry

1 answer:

wlad13 [49]3 years ago

6 0

Answer:

115g/mol

Explanation:

To get the molar mass, we know that the it is equal to the mass divided by the number of moles. We have the mass but we do not have the number of moles.

We get this by working through the solution information. Firstly, we need to know the number of moles in 750ml for a molarity of 0.29m

Now, since 0.29 moles is present in 1000ml, x moles will be present in 750ml

The value of x is obtained as follows:

x = (750 * 0.29)/1000 = 0.2175 moles

Now since we have the number of moles, we can then obtain the molar mass.

Molar mass = mass/number of moles = 25.0g/0.2175 = 114.94 approximately 105g/mol

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Which of the following shows a decrease in entropy?

shusha [124]

Precipitation hope this helped you

4 0

3 years ago

Please help with 3! Please give only the correct answer...

cupoosta [38]

The answer is: " 1.75 * 10 ^(-10) m " .
_________________________________________________________
Explanation:
_________________________________________________________
This very question asked for "Question Number 3 (THREE) ONLY, which is fine!
_________________________________________________________
Given: " 0.000000000175 m " ; write this in "scientific notation.
_________________________________________________________
Note: After the "first zero and the decimal point" {Note: that first zero that PRECEDES the decimal point in merely a "placeholder" and does not count as a "digit" — for our purposes} —
There are NINE (9) zeros, followed by "175"
_______________________________________________________
To write in "scientific notation", we find the integer that is written, as well, as any "trailing zeros" (if there are any—and by "trailing zeros", this means any number consecutive zeros/and starting with "the consecutive zeros" only —whether forward (i.e., "zeros following"; or backward (i.e. "zeros preceding").

In our case we have "zeros preceding"; that is a decimal point with zeros PRECEDING an "integer expression"
 (the "integer" is "175").
______________________________________________________
We then take the "integer expression" (whatever it may be: 12, 5, 30000001 ; or could be a negative value, etc.) ;

→ In our case, the "integer expression" is: "175" ;

and take the first digit (if the expression is negative, we take the negative value of that digit; if there is only ONE digit (positive or negative), then that is the digit we take ;

And write a decimal point after that first digit (unless in some cases, there is only one digit); and follow with the rest of the consecutive digits of that 'integer expression' ;

→ In our case: "175" ; becomes: " 1.75" .
__________________________________________________
Then we write: " * 10^ "
__________________________________________________
{that is "[times]"; or "multiplied by" : [10 raised exponentially to the power of ]._____________________________________________________
And to find that power, we take the "rewritten integer value (i.e. "whole number value that as been rewritten to a single digit with a decimal point"); and count the [number of "trailing zeros"; if there are any; PLUS the number of decimal places one goes] ; and that number is the value to which "10" is raised.
{If there are none, we write: " * 10⁰ " ; since "any value, raised to the "zero power", equals "1" ; so " * 10⁰ " ; is like writing: " * 1 " .

If there are "trailing zeros" AND/OR or any number of decimal places, to the "right" of this expression; the combined number of spaces to the right is:
{ the numeric value (i.e. positive number) of the power to which "10" is raised }.

Likewise, if there are "trailing zeros" AND/OR or any number of decimal places, to the "LEFT" of this expression; the combined number of spaces to the LEFT is the value of the power which "10" is raised to; is that number—which is a negative value.

In our case: we have: 0.000000000175 * 10^(-10) .

Note: The original notation was:

→ " 0.000000000175 m "

{that is: "175" [with 9 (nine) zeros to the left].}.

We rewrite the "175" ("integer expression") as:

"1.75" .
____________________________________________________
So we have:
→ " 0.000000000175 m " ;

Think of this value as:

" 0. 0000000001{pseudo-decimal point}75 m ".

And count the number of decimal spaces "backward" from the
"pseudo-decimal point" to the actual decimal; and you will see that there are "10" spaces (to the left).
______________________________________________________
Also note: We started with "9 (nine)" preceding "zeros" before the "1" ; now we are considering the "1" as an "additional digit" ;
→ "9 + 1 = 10" .
______________________________________________________
Since the decimals (and zeros) come BEFORE (precede) the "175" ; that is, to the "left" of the "175" ; the exponent to which the "10" is raised is:
"NEGATIVE TEN" { "-10" } .

So we write this value as: " 1.75 * 10^(-10) m " .

{NOTE: Do not forget the units of measurement; which are "meters" —which can be abbreviateds as: "m" .} .
______________________________________________________
The answer is: " 1.75 * 10^(-10) m " .
______________________________________________________

4 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

what is the proper name for the following alkyl side group where the main carbon chain is denoted with a squiggly line?

chubhunter [2.5K]

The proper name for the following alkyl side group where the main carbon chain is denoted with a squiggly line is isopropyl.

In natural chemistry, an alkyl substituent is an alkane missing one hydrogen. The term alkyl is intentionally unspecific to include many viable substitutions. An acyclic alkyl has the overall formulation of CₙH₂ₙ₊₁.

An alkyl is a purposeful institution of an organic chemical that includes only carbon and hydrogen atoms, that are organized in a chain. Examples include methyl CH3 (derived from methane) and butyl C2H5 (derived from butane). they may be now not located on their own however are discovered attached to different hydrocarbons.

what is an alkyl group? Alkyl group is shaped through putting off a hydrogen atom from the molecule of alkane. Alkanes are quite regularly represented as R-H and here R stands for alkyl group. the overall method of the alkyl organization is CₙH₂ₙ₊₁. The smallest alkyl organization is CH3 referred to as methyl.

Learn more about alkyl group here:- brainly.com/question/14272539

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4 0

2 years ago

What volume of carbon dioxide gas can be collected from

alisha [4.7K]

Answer:

1.22 L of carbon dioxide gas

Explanation:

The reaction that takes place is:

CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O

First we determine which reactant is limiting:

Calcium carbonate ⇒ 10.0 g CaCO₃ ÷ 100 g/mol = 0.10 mol CaCO₃

Hydrochloric acid ⇒ 0.100 L * 0.50 M = 0.05 mol HCl

So HCl is the limiting reactant.

Now we calculate the moles of CO₂ produced:

0.05 mol HCl * $\frac{1molCO_{2}}{1molHCl}$ = 0.05 mol CO₂

Finally we use PV=nRT to calculate the volume:

P = 1 atm

n = 0.05 mol

T = 25 °C ⇒ 25 + 273.16 = 298.16 K

1 atm * V = 0.05 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K

V = 1.22 L

7 0

3 years ago