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Artist 52 [7]
3 years ago
15

What factors help veins push blood back to the heart

Chemistry
1 answer:
Vladimir79 [104]3 years ago
5 0
Contractions of the skeletal muscles
As it puts pressure and moves the blood along
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If an object is travelling 25.0 meters/second, how far will it travel in 45 minutes?
Misha Larkins [42]
  • Speed=25m/s
  • Time=45min=(45×60)s=2700s

\\ \rm\longmapsto Distance=Speed(Time)

\\ \rm\longmapsto Distance=25(2700)

\\ \rm\longmapsto Distance=67500m

4 0
2 years ago
A 10.50 gram sample of a compound is decomposed to yield 3.40 g Na, 2.37 g S, and 4.73 g O. What is the mass percentage of each
9966 [12]

Answer:

Na = 32.4% , % S = 22.6% and %O = 45.0%

Explanation:

% Na = 3.4/10.5. × 100%

= 32.4%

%S = 2.37/10.5 × 100%

= 22.6%

% O= 4.73/10.5 × 100%

= 45.0%xplanation:

6 0
2 years ago
A 25.0 ml sample of 0.723 m hclo4 is titrated with a 0.27303 m koh solution. the h3o+ concentration after the addition of 66.2 m
tia_tia [17]
This doesn't need an ICE chart. Both will fully dissociate in water.

Assume HClO4 and KOH reacts with one another. All you need to do is determine how much HClO4 will remain after the reaction. Calculate pH.

Step 1:

write out balanced equation for the reaction

HClO4+KOH ⇔ KClO4 + H2O

the ratio of HClO4 to KOH is going to be 1:1. Each mole of KOH we add will fully react with 1 mole of HClO4

Step 2:

Determining the number of moles present in HClO4 and KOH

Use the molar concentration and the volume for each:
25 mL of 0.723 M HClO4

Covert volume from mL into L:
25 mL * 1L/1000mL = 0.025 L

Remember:

M = moles/L so we have 0.025 L of 0.723 moles/L HClO4

Multiply the volume in L by the molar concentration to get:

0.025L x 0.723mol/L = 0.0181 moles HClO4.

Add 66.2 mL KOH with conc.=0.273M
66.2mL*1L/1000mL = .0662 L
.0662L x 0.273mol/L = 0.0181 moles KOH

Step 3:

Determine how much HClO4 remains after reacting with the KOH.

Since both reactants fully dissociate and are used in a 1:1 ratio, we just subtract the number of moles of KOH from the number of moles of HClO4:

moles HClO4 = 0.0181; moles KOH = 0.0181, so 0.0181-0.0181 = 0

This means all of the HClO4 is used up in the reaction.

If all of the acid is fully reacted with the base, the pH will be neutral = 7.

Determine the H3O+ concentration:

pH = -log[H3O+]; [H3O+] = 10-pH = 10-7

The correct answer is 1.0x10-7.
3 0
3 years ago
Read 2 more answers
2. Soil is made of <br> A. air<br> B. rocks<br> C. decomposed leaves<br> D. all of the above
Irina18 [472]

Answer:

B

Explanation:

<h2>Soil is a mixture of </h2><h2 /><h2>tiny particles of rock,</h2>

<h2>dead plants and animals </h2>

<h2>,air and water.</h2>
3 0
3 years ago
Read 2 more answers
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.

<u>Answer:</u> The pH of the solution is 10.4

<u>Explanation:</u>

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^-

<u>Initial:</u>           0.39

<u>At eqllm:</u>      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
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