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3 years ago

What is true about ionic compounds?

Chemistry

1 answer:

bezimeni [28]3 years ago

6 0

It’s D I am pretty sure.

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A brick has dimensions of 15 cm x 6.0 cm x 12 cm. What is the volume of the brick in cubi meters?

Likurg_2 [28]

Convert all of the units from centimeters to meters by moving the decimal point over to the left two digits.

0.15 m x 0.06 m x 0.12 m

Volume = length x width x height
= 0.15 x 0.06 x 0.12 = 0.00108 m^3

6 0

3 years ago

What is the answer to the net ionic equation FeO(s)+2HClO4(aq)--> Fe(ClO4)2 (aq)+ H2O(l)

yan [13]

Feo + 2H = H2O + Fe + 2 + CIO4-
this is ur answer. .
mrk me as brainlist

8 0

3 years ago

Read 2 more answers

Several scientists from different countries are asked to examine the results of an experiment before a journal will print it. Wh

maxonik [38]

Answer:

Peer Review

Explanation:

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2 years ago

Read 2 more answers

Once an enzymatic reaction is completed, the enzyme releases what?

sleet_krkn [62]

Answer:

Once an enzymatic reaction is completed, the enzyme releases substrates.

Explanation:

The enzyme will always return to its original state at the completion of the reaction. One of the important properties of enzymes is that they remain ultimately unchanged by the reactions they catalyze. After an enzyme is done catalyzing a reaction, it releases its products (substrates).

5 0

2 years ago

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )

Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)

The reaction rate of equation (1) is given by:

$rate = k*[O_{3}][NO]$ (2)

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

$rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s$

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

$\frac{\Delta[NO_{2}]}{\Delta t} = rate$

$\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}$

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!

5 0

3 years ago