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Marizza181 [45]

2 years ago

9

un estudiante de noveno año, quiere realizar varias mezclas de compuestos quimicos y tiene un frasco rotulado con la formula qui

mica : MgO ¿CUAL ES EL NOMRE DEL COMPUESTO?

2 answers:

anastassius [24]2 years ago

6 0

Answer:

el óxido de magnesio es la respuesta

olganol [36]2 years ago

6 0

Answer:

Magnesium oxide (Óxido de magnesio)

Explanation:

Mg: Magnesium (Magnesio)

O: Oxygen (Oxígeno)

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A sample of neon gas occupies 105 L at 27°C under a pressure of

Viefleur [7K]

Answer: Volume occupied by given neon sample at standard condition is 123.84 L.

Explanation:

Given: $V_{1}$ = 105 L, $T_{1} = 27^{o}C = (27 + 273) K = 300 K$ , $P_{1}$ = 985 torr

At standard conditions,

$T_{2}$ = 273 K, $P_{2}$ = 760 K, $V_{2}$ = ?

Formula used to calculate the volume is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{985 torr \times 105 L}{300 K} = \frac{760 torr \times V_{2}}{273 K}\\V_{2} = \frac{94116.75}{760} L\\= 123.84 L$

Thus, we can conclude that volume occupied by given neon sample at standard condition is 123.84 L.

8 0

2 years ago

If 125 grams of Cu reacts 75 grams of HNO3, how many grams of Cu(NO3) form?

melisa1 [442]

Answer: 37.5grams of Cu(NO3)2

Cu(1mol) + 2HNO3(2mol) —> Cu(NO3)2 + H2

<em>125 grams of Cu(1mol) reacts with 75 grams of HNO3(2mol)</em>

<em><u>HNO3 is the limiting substance, therefore, 75 grams is the limiting quantity.</u></em>

<em>Therefore, 2mol of HNO3 forms 1mol of Cu(NO3)2</em>

<em>75 grams of HNO3 forms...75grams x 1mol/2mol = 37.5 grams of Cu(NO3)2</em>

3 0

2 years ago

Which property of waves is the distance between one point on a wave and that same point on the wave next to it?

mr Goodwill [35]

The answer is B, wavelength

4 0

3 years ago

Read 2 more answers

A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa

Elina [12.6K]

<u>Answer:</u> The identity of the unknown acid is butanoic acid or ascorbic acid.

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of NaOH solution = 0.570 M

Volume of solution = 39.55 mL

Putting values in above equation, we get:

$0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol$

The chemical equation for the reaction of NaOH and monoprotic acid follows:

$NaOH+HX\rightarrow NaX+H_2O$

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

So, moles of monoprotic acid = 0.0225 moles

The chemical equation for the reaction of NaOH and diprotic acid follows:

$2NaOH+H_2X\rightarrow 2NaX+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = $\frac{0.0225}{2}=0.01125moles$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

<u>For butanoic acid:</u>

Mass of butanoic acid = 2.002 g

Molar mass of butanoic acid = 88 g/mol

Putting values in above equation, we get:

$\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol$

<u>For L-tartaric acid:</u>

Mass of L-tartaric acid = 2.002 g

Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

$\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol$

<u>For ascorbic acid:</u>

Mass of ascorbic acid = 2.002 g

Molar mass of ascorbic acid = 176 g/mol

Putting values in above equation, we get:

$\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol$

As, the number of moles of butanoic acid and ascorbic acid is equal to the number of moles of acid getting neutralized.

Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

5 0

3 years ago

Please help me if your good with science:).

stellarik [79]

D an average train goes 300 km/h or 190-220 mph

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2 years ago