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Lubov Fominskaja [6]
2 years ago
13

3.2 moles of H3PO4 to grams

Chemistry
1 answer:
Nimfa-mama [501]2 years ago
4 0

Answer:

313, 6grams of H3PO4

Explanation:

We calculate the weight of 1 mol of H3PO4:

Weight 1 mol H3PO4= (Weight H)x3+ (Weight P)+(Weight 0)x4  =1gx3+31g+16gx4

Weight 1 mol H3PO4=98 g /mol

1 mol-----98 grams H3PO4

3,2mol----x= (3,2molx 98 grams H3PO4)/ 1mol=313,6 grams H3PO4

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When the forward and reverse paths of a change occur at the same rate,
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The correct answer is option B. When the forward and reverse paths of a change occur at the same rate, <span>the system is in equilibrium specifically in dynamic equilibrium.</span> Dynamic equilibrium is the balance in a process that is continuing. It is achieved in a reaction when the forward rate of reaction and the backward rate of reaction is at the same value or equal.

8 0
3 years ago
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If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0
3 years ago
(0904) How many atoms are there in 56.2 grams of Krypton, Kr?
Luba_88 [7]

Answer:

56.2÷6.02×10^23

=9.34×10^23

Explanation:

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2 years ago
Fake answers are reported and your account will be banned
Leona [35]

Answer:

BRAINLIEST?

Explanation:

Ammonia is a typical weak base. Ammonia itself obviously doesn't contain hydroxide ions, but it reacts with water to produce ammonium ions and hydroxide ions. My findings said that ammonia is a weak base, potassium hydroxide is a strong base, vinegar is a weak acid and ethyl alcohol is a weak acid.

Vinegar and ethyl alcohol are eliminated as they are acids. The question is on bases.... Potassium hydroxide is a strong base. So we are left with ammonia, being a weak base.

A is your answer

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What coefficient is missing in C2H4 + (?)O2 → 2CO2 + 2H2O?
Daniel [21]

Answer:

3

Explanation:

you must multiply everything out till everything is equal on both sides

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2 years ago
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