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3 years ago

What volume of 10M sulfuric acid is needed to create 2L solution of 7.5M?

Chemistry

1 answer:

Anvisha [2.4K]3 years ago

8 0

Explanation: the answer to life the universe and everything is 42.

School doesnt matter

drop out like i did and join the russian mob

now i live in a four floor mansion.

Do as i say

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The limiting reactant is the chemical substance that determines the amount of product(s) that can ultimately be formed in a reac

Mariulka [41]

Explanation:

Al=10,o2=5

(Al10)2(o5)3

The answer is oxygen.

6 0

3 years ago

A student fails to clean the pipet first. after delivering the vinegar sample, the student notices a drop of vinegar clinging to

VladimirAG [237]

This should not matter because the pipet has gradations and usually more of the sample is taken up in the pipette than what is delivered into the flask the student should always rinse the container being used because they are contaminating the sample if they do not clean it out

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3 years ago

A 0.1 mm sample of human blood has approximately 6000 red blood cells. An adult typically has 5.0 L of blood. How many red blood

olga55 [171]

3.0 × 10¹¹ RBC's (or) 3E11 RBC's

Solution:

Step 1: Convert mm³ into L;

As,

1 mm³ = 1.0 × 10⁻⁶ Liters

So,

0.1 mm³ = X Liters

Solving for X,

X = (0.1 mm³ × 1.0 × 10⁻⁶ Liters) ÷ 1 mm³

X = 1.0 × 10⁻⁷ Liters

Step 2: Calculate No. of RBC's in 5 Liter Blood:

As given

1.0 × 10⁻⁷ Liters Blood contains = 6000 RBC's

So,

5.0 Liters of Blood will contain = X RBC's

Solving for X,

X = (5.0 Liters × 6000 RBC's) ÷ 1.0 × 10⁻⁷ Liters

X = 3.0 × 10¹¹ RBC's

Or,

X = 3E11 RBC's

5 0

3 years ago

Which would most likely cause a person to produce antibodies?

rosijanka [135]

Answer:

receiving a vaccination

Explanation:

7 0

4 years ago

Read 2 more answers

7. NH2CO2NH4(s) when heated to 450 K undergoes the following reaction to produce a system which reaches equilibrium: NH2CO2NH4(s

Taya2010 [7]

Answer:

Value of equilibrium constant is 0.0888

Explanation:

Both $NH_{3}$ and $CO_{2}$ are gaseous. Hence equilibrium constant depends upon partial pressures of $NH_{3}$ and $CO_{2}$ .

Initially no $NH_{3}$ and $CO_{2}$ were present.

Hence mole fraction of $NH_{3}$ and $CO_{2}$ at equilibrium can be calculated from coefficient of $NH_{3}$ and $CO_{2}$ in balanced equation.

Mole fraction of $NH_{3}$ = (number of moles of $NH_{3}$ )/(total number of moles of $NH_{3}$ and $CO_{2}$ ) = $\frac{2moles}{(2+1)moles}=\frac{2}{3}$

Mole fraction of $CO_{2}$ = (number of moles of $CO_{2}$ )/(total number of moles of $NH_{3}$ and $CO_{2}$ ) = $\frac{1moles}{(2+1)moles}=\frac{1}{3}$

Let's assume both $CO_{2}$ and $NH_{3}$ behaves ideally.

Therefore partial pressure of $NH_{3}$ , $P_{NH_{3}}= x_{NH_{3}}.P_{total}$ and P_{CO_{2}}= x_{CO_{2}}.P_{total}

Where x represents mole fraction

So, $P_{NH_{3}}=\frac{2}{3}\times 0.843atm=0.562atm$

$P_{CO_{2}}=\frac{1}{3}\times 0.843atm=0.281atm$

So, $K_{p}=P_{NH_{3}}^{2}.P_{CO_{2}}=(0.562)^{2}\times 0.281=0.0888$

4 0

3 years ago