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3 years ago

What must occur ??????

Chemistry

1 answer:

Vsevolod [243]3 years ago

7 0

Air masses form fronts when B, they collide with each other

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How to balance lead and silver acetate yields lead (||) acetate and silver

arsen [322]

The given elements put into an equation using their symbols are as follows:
Pb + $AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + Ag

Since there are 2 Pb on the right side of the equation, you would change the coefficient of Pb on the left side to 2:
2Pb + $AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + Ag

Since there are 2 Acetate on the right side of the equation, you would change the coefficient of Silver Acetate on the left side to 2:
2Pb + $2AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + Ag

Now there are 2 Silver on the left side, so you change the coefficient of Silver on the right side to 2:
2Pb + $2AgC_{2}H_{3}O_{2}$ = $Pb(II){(C_{2}H_{3}O_{2})_2}$ + 2Ag

That is your final equation

The coefficients are 2 + 2 = 1 + 2

3 0

2 years ago

The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?

anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

$K_a\times K_b=K_w$

$K_b$ for sodium formate is $K_b=\frac {K_w}{K_a}$

Given that:

$K_a$ of formic acid = $1.8\times 10^{-4}$

And, $K_w=10^{-14}$

So,

$K_b=\frac {10^{-14}}{1.8\times 10^{-4}}$

$K_b=5.5556\times 10^{-11}$

Concentration = 0.35 M

HCOONa ⇒ Na⁺ + HCOO⁻

Consider the ICE take for the formate ion as:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻

At t=0 0.35 - -

At t =equilibrium (0.35-x) x x

The expression for dissociation constant of sodium formate is:

$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵ M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

5 0

3 years ago

Which statement is true about a pseudoscientific?

Alex17521 [72]

Answer:

Wereare they answer choices

Explanation:

7 0

3 years ago

Read 2 more answers

How many moles of calcium carbonate-CaCO3 = 4.15 g

marin [14]

Answer:

Number of moles = 0.042 mol

Explanation:

Given data:

Number of moles = ?

Mass of calcium carbonate = ?

Solution:

Formula:

Number of moles = mass/ molar mass

now we will calculate the molar mass of calcium carbonate.

atomic mass of Ca = 40 amu

atomic mass of C = 12 amu

atomic mass of O = 16 amu

CaCO₃ = 40 + 12+ 3×16

CaCO₃ = 40 + 12+48

CaCO₃ = 100 g/mol

Now we will calculate the number of moles.

Number of moles = 4.15 g / 100 g/mol

Number of moles = 0.042 mol

3 0

3 years ago

Calculate the volume of 2.65 moles of CO2 gas at STP.

Harman [31]

Answer:

The volume is 59, 3 liters. See the explanation below, please

Explanation:

STP conditions (standard) correspond to 273K of temperature and 1 atm of pressure. These values are used and the volume is calculated, according to the formula:

PV = nRT

1 atm x V= 2, 65 moles x 0, 082 l atm/K mol x 272 K

V= 2, 65 moles x 0, 082 l atm/K mol x 272 K/1 atm = 59, 3 liters

6 0

3 years ago