All categories

3 years ago

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?

You might be interested in

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]

Answer: The equilibrium concentration of bromine gas is 0.00135 M

Explanation:

We are given:

Initial concentration of chlorine gas = 0.0300 M

Initial concentration of bromine monochlorine = 0.0200 M

For the given chemical equation:

$2BrCl\rightleftharpoons Br_2+Cl_2$

Initial: 0.02 0.03

At eqllm: 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

Pls pls pls pls pls pls pls pls pls pls pls help help help help help help help help help help

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

Gunshot spatter

4 0

3 years ago

Read 2 more answers

HELP!!!! The freezing of methane is an exothermic change. What best describes the temperature conditions that are likely to make

vazorg [7]

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

Explanation:

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