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2 years ago

What is the maximum cold holding temperature allowed for deli meat?.

Chemistry

1 answer:

Mandarinka [93]2 years ago

6 0

40°F or less for deli to still be good

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At which instance contain 1 molecule of substances from following?

Black_prince [1.1K]

Answer:

can I have brainleis answer pls I'm new

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2 years ago

How many atoms are in a sample of 1.83 moles of potassium (K) atoms? Please explain the conversions to me. Thank you!

IRINA_888 [86]

1 mole K ------------- 6.02x10²³ atoms
1.83 moles K ------ ?? atoms

1.83 x (6.02x10²³) / 1 =

1.101x10²⁴ atoms of K

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3 years ago

What is the balance equation for HgO(s)- Hg(l)+O2(g)

artcher [175]

The balanced equation is 2HgO --> 2Hg + O2

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3 years ago

Hi ! Could someone help me out with this last part to my chemistry practice ? Thank you !

skelet666 [1.2K]

After 100years, sample is 250g

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7 0

3 years ago

Calculate the ph at of a solution of sodium hypochlorite . note that hypochlorous acid is a weak acid with a of . round your ans

loris [4]

The question is incomplete, here is the complete question:

Calculate the pH at 25°C of a 0.39 M solution of sodium hypochlorite NaClO. Note that hypochlorous acid HClO is a weak acid with a pKa of 7.50. Round your answer to 1 decimal place.

Answer: The pH of the solution is 10.4

Explanation:

We are given:

Molarity of sodium hypochlorite = 0.39 M

$pK_a$ of HClO = 7.50

We know that:

$pK_a=-\log K_a$

$K_a$ of HClO = $10^{-7.50}=3.16\times 10^{-8}$

To calculate the base dissociation constant for the given acid dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant = $3.16\times 10^{-8}$

$K_b$ = Base dissociation constant

Putting values in above equation, we get:

$10^{-14}=3.16\times 10^{-8}\times K_b\\\\K_b=\frac{10^{-14}}{3.16\times 10^{-8}}=3.16\times 10^{-7}$

The chemical equation for the reaction of hypochlorite ion with water follows:

$ClO^-+H_2O\rightarrow HClO+OH^-$

Initial: 0.39

At eqllm: 0.39-x x x

The expression of $K_b$ for above equation follows:

$K_b=\frac{[HClO][OH^-]}{[ClO^-]}$

Putting values in above equation, we get:

$3.16\times 10^{-7}=\frac{x\times x}{(0.39-x)}\\\\x=-0.00035,0.00035$

Neglecting the negative value of 'x' because concentration cannot be negative

To calculate the pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=0.00035M$

Putting values in above equation, we get:

$pOH=-\log (0.00035)=3.6$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-3.6=10.4$

Hence, the pH of the solution is 10.4

3 0

3 years ago