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3 years ago

Please help 30 points on this

Chemistry

2 answers:

Usimov [2.4K]3 years ago

8 0

A. Strong & Weak nuclear forces which are attractive. And electromagnetic.
b. Because the more electronegative atom really wants to complete it's valence shell, so it either covalently, or non-covalently bonds to the other atom.
c. Ummm, ask google? Well, it's kind of logical as well, but the part that Coulomb's law plays into it - I do not know.

Vinvika [58]3 years ago

3 0

Answer:

Explanation:

lmmmmmmmmmmmmmmmmmma0000000000000000000000000000

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Object A has a higher temperature than Object B. Which statement must be ture?

harkovskaia [24]

Object A hope this helps

God Bless

4 0

3 years ago

Read 2 more answers

4. A gas has a temperature of 14 degrees C and volume of 4.5 liters. If the temperature is raised to

nlexa [21]

Answer:

v2=4.735L

Explanation:

using charles law

v1\t1 = v2\t2

v1= 4.5L

t1= 14+273= 287K ( temperature has to be in kelvins)

t2=29+273= 302K

v2=?

by substitution;

4.5\287 = v2\302

v2=(302 x 4.5)\287

v2=4.735L

7 0

3 years ago

The formation constant* of [M(CN) 4 ]2− is 7.70 × 10 16 , where M is a generic metal. A 0.150 mole quantity of M(NO3)2 is added

iogann1982 [59]

Answer:

0 M.

Explanation:

Hello,

In this case, the undergoing reaction is:

$M(NO_3)_2+NaCN\leftrightarrow [M(CN)_4]^{-2}+NaNO_3$

Nonetheless, it only matters the reaction forming the given complex:

$M^{+2}+4CN^-\leftrightarrow [M(CN)_4]^{-2}$

In such a way, the formation constant turns out:

$K_F=\frac{[[M(CN)_4]^{-2}]_{eq}}{[M^{+2}]_{eq}[CN^{-}]_{eq}^4}$

Now, one could assume that the initial concentrations of the ions equals the original compounds concentrations:

$[M^{+2}]_0=0.150M;[CN^-]_0=0.820M$

In such a way, we modify the formation constant in terms of the change $x$ due to the reaction progress:

$K_F=\frac{x}{(0.150-x)(0.820-x)^4}=7.70x10^{16}$

Now, solving for $x$ :

$x_1=0.15M\\x_2=0.82M$

The feasible solution is 0.15M which will lead to an equilibrium concentration of M⁺² of 0M

$[M^{+2}]_{eq}=0.15M-0.15M=0M$

This fact has sense since the formation constant is very large.

Best regards.

3 0

3 years ago

An Erlenmeyer flask containing 20.0 mL of sulfuric acid of an unknown concentration was titrated with exactly 15.0 mL of 0.25 M

MrRa [10]

Answer:

0.09375M

Explanation:

There are two methods of going about this, either we use dilution formula (easiest and fastest) or we solve through molarity.

Using dilution formula,

C2 (H2SO4) = ?

C1 (NaOH) = 0.25M

V2 (H2SO4) = 20cm³

V1 (NaOH) = 15cm³

However we can solve using molarity method

Equation of reaction =

2NaOH + H2SO4 ====》 Na2SO4 + 2H2O

O.25M of NaOH = 1000cm³

X moles = 15cm³

X = (0.25 * 15) / 1000

X = 0.00375 moles is present in 15cm³ of NaOH

From equation of reaction,

2 moles of NaOH requires 1 mole of H2SO4

Therefore

0.00375 / 2 = 0.001875 moles is present in H2SO4

From the reaction,

0.00187 moles of H2SO4 = 20 cm³

X moles = 1000cm³

X = (0.00187*1000) / 20 = 0.09375M

8 0

3 years ago

Sarah measures out 151 grams of SO2. How many moles is this? Express your answer to three significant figures.

Anna11 [10]

Answer:

$\boxed {\boxed {\sf 2.36 \ mol \ SO_2}}$

Explanation:

We are asked to convert grams to moles. We will use the molar mass and dimensional analysis to perform this conversion.

<h3>1. Molar Mass</h3>

The molar mass is the mass of 1 mole of a substance. These values are found on the Periodic Table because they are equivalent to the atomic masses, but the units are grams per mole instead.

We are given a mass of sulfur dioxide (SO₂). Look up the molar masses of the individual elements.

Sulfur (S): 32.07 g/mol
Oxygen (O): 15.999 g/mol

Notice that the formula of the compound contains a subscript. The subscript after O means there are 2 moles of oxygen in 1 mole of sulfur dioxide. We must multiply oxygen's molar mass before adding sulfur's.

O₂: 15.999 * 2 = 31.998 g/mol
SO₂= 32.07 + 31.998 = 64.068 g/mol

<h3>2. Convert Grams to Moles </h3>

Now we will use dimensional analysis to convert grams to moles. From the molar mass, we know there are 64.068 grams of sulfur dioxide per mole, so we can set up a ratio.

$\frac {64.068 \ g \ SO_2} {1 \ mol \ SO_2}$

We are converting 151 grams to moles, so we multiply by this value.

$151 \ g \ SO_2 *\frac {64.068 \ g \ SO_2} {1 \ mol \ SO_2}$

Flip the ratio so the units of grams of sulfur dioxide cancel.

$151 \ g \ SO_2 *\frac {1 \ mol \ SO_2}{64.068 \ g \ SO_2}$

$151 *\frac {1 \ mol \ SO_2}{64.068 }$

$\frac {151}{64.068 } \ mol \ SO_2$

$2.356870825 \ mol \ SO_2$

<h3>3. Round </h3>

The original measurement of grams (151) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 6 in the thousandth place tells us to round the 5 in the hundredth up to a 6.

$2.36 \ mol \ SO_2$

151 grams of sulfur dioxide is approximately <u>2.36 moles of sulfur dioxide.</u>

8 0

3 years ago