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3 years ago

ZnSO4 + + LINO, → ---- Zn(NO3)2 + Li,SO4 Balanced equation

Chemistry

1 answer:

Tanzania [10]3 years ago

4 0

Answer:

ZnSO4 + 2LiNO3 → Zn(NO3)2 + Li2SO4

Explanation:

There's many resources on web that can assist you with this concept:

https://en.intl.chemicalaid.com/tools/equationbalancer.php

https://www.webqc.org/balance.php

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When the following two solutions are mixed:

Gnom [1K]

Answer:

the two spectator ions are; K(+) and NO3(-)

Explanation:

First off, let's write out the balanced chemical equation for the reaction;

3K2CO3(aq) +2Fe(NO3)3(aq) ----> 6KNO3(aq) + Fe2(CO3)3(s)

In order to identify which ions are spectators, we have to break the equation down to an ionic equation. This is done by splitting all aqueous compounds into ions while leaving the solids, liquids as they are.

We have;

K(+) + CO3(2-) + Fe(3+) + NO3(-) ---> K(+) + NO3(-) + Fe2(CO3)3(s)

Spectators ions are pretty much those ions that do not undergo a change in the reaction. Spectator ions always have the same number of moles and charge in both sides of the reaction.

Upon observing the ionic equation, we can tell that the two spectator ions are; K(+) and NO3(-)

3 0

3 years ago

Debbie was doing a science lab, and she was instructed to decide whether a piece of plastic would float or sink without actually

miv72 [106K]

''Find the density of the plastic and compare it to the density of water''

7 0

3 years ago

Read 2 more answers

You may want to reference (Page) Section 12.4 while completing this problem. Enter the balanced chemical equation for the comple

alexandr1967 [171]

Answer:

See explanation

Explanation:

In this question, we have to remember that in all combustion reactions we will have Oxygen as reactive ( $O_2$ ) and the products are Carbon dioxide ( $CO_2$ ) and Water ( $H_2O$ ). Additionally, for the states, we will have (l) for the liquid state and (g) for the gas state. So, we can analyze each reaction:

Part A.

The formula for nonane is $C_9H_2_0$ , with this in mind we can write the combustion reaction:

$C_9H_2_0_(_I_)~
+~
O_2~->~CO_2_(_g_)~
+~H_2O_(_g_)$

When we balance the reaction we will obtain:

$C_9H_2_0_(_I_)~
+~14 O_2_(_g_)~->~9CO_2_(_g_)~
+~10H_2O_(_g_)$

Part B.

The formula for 2-methylbutane is $C_5H_1_2$ , with this in mind we can write the combustion reaction:

$C5H12_(_I_)~
+~O_2_(_g_)~->~CO_2_(_g_)~
+~H2O_(_g_)$

When we balance the reaction we will obtain:

$C5H12_(_I_)~
+~8O_2_(_g_)~->~5 CO_2_(_g_)~
+~6 H2O_(_g_)$

Part C.

The formula for 3-ethyltoluene is $C_9H_1_2$ , with this in mind we can write the combustion reaction:

$C_9H_1_2_(_I_)~+~O_2_(_g_)~->~CO_2_(_g_)~+~H2O_(_g_)$

When we balance the reaction we will obtain:

$C9H12_(_I_)~
+~12O_2_(_g_)~->~9CO_2_(_g_)~
+~6H2O_(_g_)$

See figure 1 for further explanations.

I hope it helps!

8 0

3 years ago

The half-life of tritium, or hydrogen-3, is 12.32 years. After about 24.6 years, how much of a sample of tritium will remain unc

scoundrel [369]

Answer: The correct option is B.

Explanation: This is an example of radioactive decay and all the radioactive decay processes follow First order of kinetics.

Expression for the half life of first order kinetics is:

$t_{1/2}=\frac{0.693}{k}$

We are given:

$t_{1/2}=12.32years$

Putting in above equation, we get:

$12.32=\frac{0.693}{k}\\k=0.05625year^{-1}$

Expression to calculate the amount of sample which is unchanged is:

$N=N_oe^{-kt}$

where,

N = Amount left after time t

$N_o$ = Initial amount

k = Rate constant

t = time period

Putting value of k = 0.05625 and t = 24.6 in above equation, we get:

$N=N_oe^{-0.05625\times 24.6}$

$\frac{N}{N_o}=0.25$

The above fraction is the amount of sample unchanged and that is equal to $\frac{1}{4}$

Hence, the correct option is B.

8 0

3 years ago

Read 2 more answers

Se realiza una mezcla de minerales de Cu y Fe: 20 kg FeS2 (pirita), 70 kg de Fe2O3 (hemetita) 15 kg de CuFe2 (calcopirita) y 90

Artemon [7]

Answer:

34.78% Fe

39.66% Cu

5.48% S

20.07% O

Explanation:

Para resolver esta pregunta debemos hallar la masa de cada átomo en cada mineral. Así, podremos hallar el porcentaje de cada átomo:

Pirita (Fe: 55.845g/mol; S: 32.065g/mol; FeS2: 119.975g/mol)

Masa Fe:

20kg FeS2 * (1*55.845g/mol / 119.975g/mol) = 9.31kg Fe

Masa S:

20kg FeS2 * (2*32.065g/mol / 119.975g/mol) = 10.69kg S

Hemetita (Fe: 55.845g/mol; O: 16g/mol; Fe2O3: 159.688g/mol)

Masa Fe:

70kg Fe2O3 * (2*55.845g/mol / 159.688g/mol) = 48.96kg Fe

Masa O:

70kg Fe2O3 * (3*16g/mol / 159.688g/mol) = 21.04kg O

Calcopirita (Fe: 55.845g/mol; Cu: 63.546g/mol; CuFe2: 175.236 g/mol)

Masa Fe:

15kg CuFe2 * (2*55.845g/mol / 175.236 g/mol) = 9.56kg Fe

Masa Cu:

15kg CuFe2 * (1*63.546g/mol / 175.236 g/mol) = 5.44kg Cu

Tenorita (O: 16g/mol; Cu: 63.546g/mol; CuO: 79.545 g/mol)

Masa O:

90kg CuO * (1*16g/mol / 79.545 g/mol) = 18.10kg O

Masa Cu:

90kg CuO * (1*63.546g/mol / 79.545 g/mol) = 71.90kg Cu

Masa Total: 20kg + 70kg + 15kg + 90kg = 195kg

Porcentaje Hierro:

9.31kg Fe + 48.96kg Fe + 9.56kg Fe / 195kg * 100 =

34.78% Fe

Porcentaje Cobre:

5.44kg Cu + 71.90kg Cu / 195kg * 100 =

39.66% Cu

Porcentaje Azufre:

10.69kg S / 195kg * 100 =

5.48% S

Porcentaje Oxígeno:

21.04kg O + 18.10kg O/ 195kg * 100 =

20.07% O

8 0

3 years ago