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18 days ago

Persamaan setara pada reaksi besi dengan asam klorida membentuk besi (II) klorida dan gas hidrogen

Chemistry

1 answer:

VMariaS [2.8K]18 days ago

8 0

<span>Reaksi antara besi dan asam klorida menghasilkan besi (II) klorida serta gas hidrogen.</span>

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which intensive physical property is observed when droppings of a person seated inside a closed room has able to reach a person

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Answer:

The correct answer is "Speed".

Explanation:

An intensive or individualized physical property is identified when "speed" is observed as the excretion of an individual in a confined area, capable of reaching someone one meter away after sneezing or coughing.
This measure is represented in the unit of "meter per second", indicating its intensive nature.

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1 month ago

A given sample of caffeine, C8H10N4O2, has 6.47 x 1022

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8.08 × atoms of hydrogen.

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23 days ago

6.74 g of the monoprotic acid KHP (MW = 204.2 g/mol) is dissolved into water. The sample is titrated with a 0.703 M solution of

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Answer:

The volume of calcium hydroxide solution utilized is 0.0235 mL.

Explanation:

$2KHP+Ca(OH)_2\rightarrow 2H_2O+Ca(KP)_2$

Moles of KHP = $\frac{6.74 g}{204.2 g/mol}=0.0330 mol$

In accordance with the reaction, 2 moles of KHP react with 1 mole of calcium hydroxide, thus 0.0330 moles of KHP will react with;

$\frac{1}{2}\times 0.0330 mol=0.01650 mol$ of calcium hydroxide

The molarity of calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

$Molarity=\frac{Moles}{Volume(L)}$

$0.703 M=\frac{0.01650 M}{V}$

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The volume of the calcium hydroxide solution utilized is 0.0235 mL.

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19 days ago

In an experiment a student mixes a 50.0 mL sample of 0.100 M AgNO₃(aq) with a 50.0 mL sample of 0.100 M NaCl(aq) at 20.0°C in a

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The enthalpy change associated with the precipitation reaction is 84 kJ/mole

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The chemical equation for the reaction can be written as

AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)

To determine the enthalpy change, the following equation applies

$\Delta H =\frac{Q}{n}$

To calculate the heat (Q):

$Q=m*C*\Delta T=(100g)*(4.2 J/g*^\circ C)*(21^\circ C-20^\circ C)\\\\Q=420J$

Next, we need to calculate the number of moles involved in the reaction (n):

$n=[AgNO_3]*v(L)=(0.1M)*(0.05L)=0.005moles$

With these two values, we can substitute them into the first equation:

$\Delta H= \frac{420J}{0.005moles}=84000J/mole=84kJ/mole$

Have a great day!

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1 month ago

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Alekssandra [2904]

In this case, to find the grams of sodium chloride starting from its molecules, the first step is to determine the moles of sodium chloride by utilizing Avogadro's number. After that, we can obtain grams directly using the molar mass of sodium chloride (58.45 g/mol).

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16 days ago