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3 years ago

Find the mass of an object that has a density of 1.5 g/cm3 and has a volume of 8cm3

Chemistry

1 answer:

zubka84 [21]3 years ago

7 0

Answer:

density= 1.5 g/cm3

volume= 8cm 3

mass = density×volume

mass= 1.5×8

= 12 g

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Which is the larger atom kr Or As

nasty-shy [4]

Answer:

I think As is larger

Explanation:

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3 years ago

What is the molarity of 60.0 grams of NaOH dissolved in 750 milliliters of water?

Bumek [7]

Answer:

The correct answer option is B) 2.0 M.

Explanation:

We are given the number of grams of NaOH (Sodium Chloride) which are dissolved in 750 milliliters of water and we are to find its molarity.

We know the formula of molarity:

Molarity = (mass * 1000) / (volume * molecular mass)

Volume = 750 ml = 750 cm $^3$

Molecular mass = 40

Mass = 60 grams

Substituting these values in the above formula:

Molarity = $\frac{(60*1000)}{750*40}$ = 2.0 M

8 0

3 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

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3 years ago

Elemental Li and Na are prepared by electrolysis of a molten salt, whereas K, Rb, and Cs are prepared by chemical reduction.(c)

pishuonlain [190]

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Learn more about atomic radius here:

brainly.com/question/13607061

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2 years ago

How many moles of mercury, Hg, are there in 1.30 x 10^7 atoms of murcury? PLEASE HELP, IT'S URGENT! Thank you! (:

vodka [1.7K]

1 mole Hg ---------------- 6.02x10²³ atoms
?? ------------------------- 1.30 x10⁷ atoms

1.30x10⁷ x 1 / 6.02x10²³ =

= 1.30x10⁷ / 6.02x10²³ => 2.159x10⁻¹⁷ moles

hope this helps!

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3 years ago