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3 years ago

6

Use the periodic table to identify the number of core electrons and the number of valence electrons in each case below.Magnesium

(Mg): 1s22s22p63s2

2 answers:

amm18123 years ago

3 0

Answer:

1s1

2s2/2p6

3s2⇨⇨ period 3 - group 2

laila [671]3 years ago

3 0

There’s 2 valence electrons in Mg

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Which of the following obtain their energy from the organisms they eat?

Zepler [3.9K]

The white-tailed deer. (D)

The rest are plants, which make their own energy from photosynthesis.

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3 years ago

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En un recipiente cerrado y rígido se introdujo una mezcla gaseosa a cierta temperatura y las presiones parciales de cada gas son

valkas [14]

Answer:

Qp > Kp, por lo tanto, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.

Explanation:

Paso 1: Escribir la ecuación balanceada

BrF₃ (g) ⇌ BrF(g) + F₂(g) Kp(T) = 64,0

Paso 2: Calcular el cociente de reacción (Qp)

Qp = pBrF × pF₂ / pBrF₃

Qp = 1,50 × 2,00 / 0,0150 = 200

Paso 3: Sacar una conclusión

Dado que Qp > Kp, la reacción se desplazará hacia la izquierda para alcanzar el equilibrio, es decir, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.

7 0

3 years ago

In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL +

Volgvan

<u>Answer:</u> The pH of the solution is 1.136

<u>Explanation:</u>

To calculate the moles from molarity, we use the equation:

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

<u>For ammonia:</u>

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol$

<u>For nitric acid:</u>

Molarity of nitric acid = 0.3838 M

Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

$0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol$

After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol

For the reaction of ammonia with nitric acid, the equation follows:

$NH_3+HNO_3\rightarrow NH_4NO_3$

At $t=0$ 0.0178 0.022

Completion 0 0.0042 0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

$pH=-\log[H^+]$

where,

$[H^+]=\frac{0.0042mol}{0.05741L}=0.0731M$

Putting values in above equation, we get:

$pH=-\log(0.0731)\\\\pH=1.136$

Hence, the pH of the solution is 1.136

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3 years ago

Which statement describes a property that is unique to metalloids?

hodyreva [135]

Answer:

Metalloids are semiconductive.

Explanation:

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3 years ago

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A star is estimated to have a mass of 2.0 x 10 ^36kg. Assuming it to be a sphere of average radius of 7.0 x 10 ^5 km. Calculate

Montano1993 [528]

Answer:

<em>a)</em> <em>1.392 x 10^6 g/cm^3</em>

<em>b) 8.69 x 10^7 lb/ft^3</em>

<em></em>

Explanation:

mass of the star m = 2.0 x 10^36 kg

radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m

The density of substance ρ = mass/volume

The volume of the star = volume of a sphere = $\frac{4}{3}\pi r^{3}$

==> V = $\frac{4}{3}*3.142*(7.0*10^8)^{3}$ = 1.437 x 10^27 m^3

density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3

in g/cm^3 = (1.392 x 10^9)/1000 = <em>1.392 x 10^6 g/cm^3</em>

in lb/ft^3 = (1.392 x 10^9)/16.018 = <em>8.69 x 10^7 lb/ft^3</em>

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3 years ago