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3 years ago

Can someone help me with the question in the image. if it is correct i will mark as brainliest

Chemistry

1 answer:

Juliette [100K]3 years ago

6 0

Answer:

If I'm right, I beleive it is the last question-

I really hope this helps ^-^

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can we stir silver nitrate solution with copper spoon? why or why not? support your answer with reason.

HACTEHA [7]

↪ Hello miss "galaxyu035" here's your answer for the question you wrote.

Answer :

No, we cannot stir silver nitrate solution with a copper spoon because , copper is more reactive than silver. Thus,it would displace silver from the silver nitrate solution forming copper nitrate.

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8 0

4 years ago

A chemist prepares a solution of aluminum sulfite by measuring out 109 umol of aluminum sulfite into a 350 mL volumetric flask a

Lelu [443]

Answer:

0.311 mmol/L

Explanation:

109 μmol = 109*10^(-6) mol

109*10^(-6) mol = 109*10^(-6) mol*(10^3 mmol/1mol) = 109*10^(-3) mmol =

=0.109 mmol

350 mL = 0.350 L

0.109 mmol/0.350 L = 0.311 mmol/L

6 0

4 years ago

What factor is most important in determining whether water is a solid liquid or gas

Usimov [2.4K]

Answer:

Temperature

Explanation:

temperatures below 0 degrees celsius will turn water into ice

Temperatures 1 to 99 degrees celsius is water in liquid form

Temperatures of over 100 degrees celsius liquid turns into gas

6 0

4 years ago

Read 2 more answers

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

What point does the ammonium ions act as a buffer solution? Please this is driving me insane.

Shalnov [3]

Answer:

Explanation:The buffer is a mixture of ammonia (NH3) and ammonium (NH4+). ... Since ammonia (NH3) is a weak base, it will have a pH above 7 and since ammonium (NH4+) is a weak acid, it will have a pH below 7. Ammonia (NH3) is a weak base and ammonium (NH4+) is a weak acid. Ammonia (NH3) is the conjugate base of ammonium (NH4+).

7 0

3 years ago