Which statement best describes how an existing theory is often affected by the development of new technology?

A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4

ddd [48]

Answer:

178.35g

Explanation:

Molarity of a solution can be calculated using the formula:

Molarity = number of moles ÷ volume

Based on the information provided in this question, molarity (M) of the solution = 1.50 M, volume = 725 mL = 725/1000 = 0.725L, n = ?

1.50 = n / 0.725

n = 1.50 × 0.725

n = 1.0875mol

Molar mass of Na3PO4

23(3) + 31 + 16(4)

= 69 + 31 + 64

= 164g/mol

Mole = mass ÷ molar mass

1.0875 = mass/164

mass = 178.35g

5 0

3 years ago

A 1.0-L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys the buffer? (a) adding 0.050 mol of HCl (b) ad

Volgvan

Answer:

(a) adding 0.050 mol of HCl

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.

In the buffer:

1.0L × (0.10 mol / L) = 0.10 moles of HF -Weak acid-

1.0L × (0.050 mol / L) = 0.050 moles of NaF -Conjugate base-

-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-

Thus:

(a) adding 0.050 mol of HCl: The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid destroying the buffer.

(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer wouldn't be destroyed.

(c) adding 0.050 mol of NaF: The addition of conjugate base doesn't destroy the buffer

3 0

3 years ago

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

Write balanced complete ionic equation for hcl(aq)+lioh(aq)→h2o(l)+licl(aq) express your answer as a chemical equation. identify

Svet_ta [14]

Explanation:

An equation is said to be balanced when the number of atoms on both reactant and product side are equal in number.

Whereas an equation where electrolytes in an aqueous solution are represented as dissociated ions is known as an ionic equation.

For example, $HCl(aq) + LiOH(aq) \rightarrow H_{2}O(l) + LiCl(aq)$ can be represented in ionic form as follows.

$H^{+} + Cl^{-} + Li^{+} + OH^{-} \rightarrow H_{2}O(l) + Li^{+} + Cl^{-}$

Now, cancelling the common ions present on both sides of the equation. The resulting, ionic equation will be as follows.

$H^{+} + OH^{-} \rightarrow H_{2}O(l)$

8 0

3 years ago

Read 2 more answers

Which of the following is most likely to form multiple (double or triple) bonds? 1. Li 2. Cl 3. N 4. H 5. F

alukav5142 [94]

Answer: Option (3) is the correct answer.

Explanation:

Atomic number of lithium is 3 and its electronic distribution is 2, 1. So, to attain stability it will loose an electron and hence, it forms a single bond.

Atomic number of chlorine is 17 and it has 7 valence electrons. Hence, in order to attain stability it will gain one electron and therefore, it forms a single bond only.

Atomic number of nitrogen is 7 and its electronic distribution is 2, 5. Therefore, to attain stability it needs to gain 3 more electrons. Hence, a nitrogen atom is able to form a triple bond and also it is able to form a double bond.

Hydrogen has atomic number 1 and it attains stability by gaining one electron. Therefore, a hydrogen atoms always forms a single bond.

Atomic number of fluorine is 9 and its electronic distribution is 2, 7. To complete its octet it needs to gain one electron. Hence, a fluorine atom always forms a single bond.

Thus, we can conclude that out of the given options nitrogen is most likely to form multiple (double or triple) bonds.

3 0

3 years ago