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3 years ago

15

Um...id.k how to do chemistry...help please?

2 answers:

Cerrena [4.2K]3 years ago

4 0

Answer:

um...id.k how to do chemistry...help please?

Explanation:

NaBr

qaws [65]3 years ago

3 0

Answer:

NaBr

Explanation:

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Is neutralizing a base, physical or chemical property?

DIA [1.3K]

Physical property

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3 years ago

Read 2 more answers

the empirical formula of a compound is NO2. it's molecular mass is 92g/mol. What is its molecular formula?

lesantik [10]

The molecular formula:
$N_x(O_2)_x$

$M_{N_x(O_2)_x}=x \times M_N + x \times 2 \times M_O \\ \\
m_N=14 \ \frac{g}{mol} \\
m_O=16 \ \frac{g}{mol} \\ M_{N_x(O_2)_x}=92 \ \frac{g}{mol} \\ \\
92=x \times 14+x \times 2 \times 16 \\
92=14x+32x \\
92=46x \\
\frac{92}{46}=x \\
x=2 \\ \\
N_2(O_2)_2 \Rightarrow N_2O_4$

The molecular formula is N₂O₄.

8 0

3 years ago

What is the correct term for the breakdown of organic sediment into phosphorous

viktelen [127]

Answer: Phosphorous Cycle

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3 years ago

a chemical spill gets into the soil and effects the plants within a forested ecosystem. explain why the top predators would be m

notka56 [123]

Well if we think about a food pyramid there are consumers that eat plants and if there are no plants to eat then those primary consumers die. Which means that the secondary consumers have no prey to eat and die as well, eventually leading to the top predators who have absolutely nothing to feed on.
They are the most affected because energy level decreases the higher up you go so the top predators need to eat more and if there is less and less to eat they die out quicker

3 0

3 years ago

Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) I2(g) ⇌ 2HI(g) is 54.3 at 430°C. Determine the init

Jlenok [28]

Answer : The initial concentration of HI and concentration of $HI$ at equilibrium is, 0.27 M and 0.386 M respectively.

Solution : Given,

Initial concentration of $H_2$ and $I_2$ = 0.11 M

Concentration of $H_2$ and $I_2$ at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

$H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$

Initially 0.11 0.11 C

At equilibrium (0.11-x) (0.11-x) (C+2x)

As we are given that:

Concentration of $H_2$ and $I_2$ at equilibrium = 0.052 M = (0.11-x)

0.11 - x = 0.052

x = 0.11 - 0.052

x = 0.058 M

The expression of $K_c$ will be,

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}$

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of $HI$ at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

3 0

4 years ago