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OleMash [197]
3 years ago
14

PLEASE HELP!! GIVING BRAINLIEST

Chemistry
1 answer:
RSB [31]3 years ago
4 0

Answer:

The answer is "0.00172172603".

Explanation:

Given:

Mass (M) = 1.60 \times  10^{-3} \ g\\\\Density (D) = 9.293 \times 10^{-1} \ \frac{g}{cm^3}\\\\Volume (V) =  ?

Formula:

\to \bold{V = \frac{M}{D}}

       = \frac{1.60 \times  10^{-3}}{9.293 \times  10^{-1}} \\\\  = \frac{1.60 \times  10^{1}}{9.293 \times  10^{3}} \\\\ = \frac{1.60 \times  10}{9.293 \times  1000} \\\\ = \frac{1.60 }{9.293 \times  100} \\\\ = \frac{1.60 }{929.3 } \\\\= 0.00172172603

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suppose you find a website that describes rusting of a nail as a change to only the physical properties of iron. would it be acc
Kaylis [27]

Answer:

The website is inaccurate.

Explanation:

Remember, a chemical change is characterized by many things, but most commonly:

  1. A change in color, temperature or state.
  2. An irreversible change
  3. One or more new substances are formed

Rusting (Iron oxide) is a redox reaction (Oxidation-Reduction Reaction) where  iron molecules come in contact with water and oxygen molecules. The iron molecules lose electrons, becoming oxidized, while oxygen gains electrons, becoming reduced.

Is rust reversable? (No)

Did the colour of the iron on the surface change? (Yes)

Did a new substance form? (Yes, Read above).

Hope I helped! Have a good day!

5 0
2 years ago
Consider the following reaction: Consider the reaction 2NO(g)+Br2(g)⇌2NOBr(g) ,Kp=28.4 at 298 K In a reaction mixture at equilib
Vitek1552 [10]

Answer:  partial pressure of NOBr is 7792 atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

2NO(g)+Br_2(g)\rightleftharpoons 2NOBr(g)

Equilibrium constant is given as:

K_{p}=\frac{[p_{NOBr}]^2}{[p_{NO}]^2\times [p_{Br_2}]^1}

28.4=\frac{[p_{NOBr}]^2}{[(119)^2\times (151)^1}

[p_{NOBr}]=7792 atm

Partial pressure of NOBr is 7792 atm

4 0
3 years ago
Empirical formula for 70.9% K and 29.1% S
aev [14]
I dont know sorry but good luck i know u can past
4 0
3 years ago
When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

3 0
3 years ago
What should you do during a titration when you notice the indicator start to indicate the approach of the equilibrium point? Add
Leni [432]

B. Add the second reactant slower.

6 0
3 years ago
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