All categories

3 years ago

PLEASE HELP

2 answers:

7 0

Since the gravitational force is directly proportional to the mass of both interacting objects, more massive objects will attract each other with a greater gravitational force. So as the mass of either object increases, the force of gravitational attraction between them also increases.

So C, I think?

prohojiy [21]3 years ago

3 0

Answer:

Mass and gravitational for e is not relart

You might be interested in

C. 49.8 grams of ki is dissolved in enough water to make 1.00 l of solution. what is the molarity?

taurus [48]

Hey There!:

Molar Mass KI => 166.003 g/mol

* number of moles:

n = mass of solute / molar mass

n = 49.8 / 166.003

n = 0.3 moles KI

Therefore:

M = n / V

M = 0.3 / 1.00

M = 0.3 mol/L

hope this helps!

7 0

3 years ago

Read 2 more answers

Helppppp asaapppppp plzzzzzz

Gala2k [10]

Answer:

Alright the very first thing you need to do is balance the equation:

2HCl + Na2CO3 -----> 2NaCl + CO2 + H2O

Now we need to find the limiting reactant by converting the volume to moles of both HCl and Na2CO3.

Volume x Concentration/molarity = moles

0.235L x 0.6 M = 0.141 moles / molar ratio of 2 = 0.0705 moles of HCl

0.094L x 0.75 M = 0.0705 moles /molar ratio of 1 = 0.0705 moles of Na2CO3

Since both of the moles are equal, it means the entire reaction is complete (while the identification of limiting reactant may seem like an unnecessary step, it's quite essential in stoichiometry, so keep an eye out) and there is no excess of any reactant.

Now we know that the product we want to calculate is aqueous so, following the law of conservation of mass, we should add both volumes together to calculate how much volume we could get for NaCl.

0.235 + 0.094 = 0.329L of NaCl

Now we apply the C1V1 = C2V2 equation using the concentration and volume of Na2CO3 because it's molar ratio is one to one to NaCl (You can also use HCL, but you have to divide their moles by 2 for the molar ratio) and the volume we just calculated for NaCl.

(0.75M) x (0.094L) = C2 x (0.329L)

Rearrange equation to solve for C2:

<u>(0.75M) x (0.094L)</u> = C2

(0.329L)

C2 = 0.214 M (Rounded)

<u>When the reaction is finished, the NaCl solution will have a molarity concentration of 0.214 M.</u>

<u></u>

<u />

7 0

2 years ago

When 33.3 grams of propane (C3H8) undergoes combustion, what is the theoretical yield of water in grams? The molar mass of p

aev [14]

C3H8+ 5 O2 --> 3 CO2 + 4 H2O
44 g. --------> 72 g
33.3 g. --------> x
$x = \frac{33.3 \times 72}{44} \\ x = 54.5 \: g$
Answer: The theoretical yield of H2O is 54.5

3 0

3 years ago

Read 2 more answers

"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"

Salsk061 [2.6K]

Answer:- The Ka for the acid is $1.53*10^-^5$ .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

$Ka = [H^+][A^-]\frac{1}{HA}$

Where, Ka is the acid ionization constant. Let's plug in the values.

$Ka = \frac{X^2}{0.25-X}$

Let's calculate the value of X first using the equation:

$pH = -log[H^+]$ [/tex]

on taking antilog ob above equation we get:

$[H^+]=10^-^p^H$

$[H^+]=10^-^2^.^7^1$

$[H^+]$ = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

$Ka=\frac{(0.00195)^2}{0.25-0.00195}$

$Ka=1.53*10^-^5$

So, the value of Ka for butyric acid is $1.53*10^-^5$ .

8 0

3 years ago

Read 2 more answers

A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. Calculate the pressure when the volume is 1.41 L and the

Vlad1618 [11]

A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

<h3>What is Combined Gas Law ?</h3>

This law combined the three gas laws that is (i) Charle's Law (ii) Gay-Lussac's Law and (iii) Boyle's law.

It is expressed as

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where,

P₁ = first pressure

P₂ = second pressure

V₁ = first volume

V₂ = second volume

T₁ = first temperature

T₂ = second temperature

Now put the values in above expression we get

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

$\frac{1.02\ atm \times 2.30\ L}{281\ K} = \frac{P_2 \times 1.41\ L}{298\ K}$

$P_{2} = \frac{1.02\ atm \times 2.30\ L \times 298\ K}{281\ K \times 1.41\ L}$

P₂ = 1.76 atm

Thus from the above conclusion we can say that A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

Learn more about the Combined gas Law here: brainly.com/question/13538773

#SPJ4

4 0

2 years ago