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Serggg [28]
3 years ago
14

The density of silver is 10.49 g/cm3. if a sample of pure silver has a volume of 12.993 cm3, what would the mass?

Chemistry
1 answer:
weqwewe [10]3 years ago
6 0
Multiply 10.49 by 12.993. that should be it. 130 grams ish?
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In a titration of 0.35 M HCl and 0.35 M NaOH, how much NaOH should be added to 45.0 ml of HCl to completely neutralize the acid?
Bezzdna [24]
It would be the same amount. So, 45 ml of NaOH is required to be added to the 45 ml of HCI to neutralize the acid fully. Here is a brief calculation:

Firstly, here is your formula: M(HCI) x V(HCI) = M(NaOh) x V(NaOH) 
With the values put in: 0.35 x 45 = 0.35 x V(NaOH) 
= 45 ml. 
There is 45 ml of V(NaOH)

Let me know if you need anything else. :)

           - Dotz
6 0
3 years ago
At 25°C, the equilibrium constant Kc for the reaction in thesolvent CCl4 2BrCl <----> Br2 + Cl2 is 0.141. If the initial c
ivolga24 [154]

<u>Answer:</u> The equilibrium concentration of bromine gas is 0.00135 M

<u>Explanation:</u>

We are given:

Initial concentration of chlorine gas = 0.0300 M

Initial concentration of bromine monochlorine = 0.0200 M

For the given chemical equation:

                   2BrCl\rightleftharpoons Br_2+Cl_2

<u>Initial:</u>          0.02               0.03

<u>At eqllm:</u>    0.02-2x     x     0.03+x

The expression of K_c for above equation follows:

K_c=\frac{[Br_2]\times [Cl_2]}{[BrCl]^2}

We are given:

K_c=0.141

Putting values in above equation, we get:

0.141=\frac{x\times (0.03+x)}{(0.02-2x)^2}\\\\x=-0.96,+0.00135

Neglecting the value of x = -0.96 because, concentration cannot be negative

So, equilibrium concentration of bromine gas = x = 0.00135 M

Hence, the equilibrium concentration of bromine gas is 0.00135 M

8 0
3 years ago
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Gwar [14]

Answer:

Explanation:

The moon would fall and earth as well

6 0
3 years ago
What are the Eight types of bloodstain patterns ?
valentina_108 [34]

Passive bloodstain

Transfer pattern

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4 0
3 years ago
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HELP!!!! The freezing of methane is an exothermic change. What best describes the temperature conditions that are likely to make
vazorg [7]

<u>Answer:</u> The correct statement is low temperature only, because entropy decreases during freezing.

<u>Explanation:</u>

The relationship between Gibb's free energy, enthalpy, entropy and temperature is given by the equation:

\Delta G=\Delta H-T\Delta S

Where,

\Delta G = change in Gibb's free energy

\Delta H = change in enthalpy

T = temperature

\Delta S = change in entropy

It is given that freezing of methane is taking place, which means that entropy is decreasing and Delta S is becoming negative. It is also given that the reaction is an exothermic reaction, this means that the \Delta H is also negative.

For a reaction to be spontaneous, \Delta G must be negative.

-ve=-ve-[T(-ve)]\\\\-ve=-ve+T

From above equations, it is visible that \Delta G will be negative only when the temperature will be low.

Hence, the correct statement is low temperature only, because entropy decreases during freezing.

8 0
3 years ago
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