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2 years ago

At what year will Sagittarius be visible from the earth’s North Pole?

1 answer:

8 0

Best visible at 9.pm during the month of August. (Sagittarius is best observed during the Northern hemisphere summer months) .. I hope this helped ?

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What is a decay series?

denpristay [2]

Answer:

Explanation:

because in a decay series a daughter nuclei may be stable or decay itself. that starts a decay series

5 0

3 years ago

Read 2 more answers

How many atoms are in each compound ?

noname [10]

Compounds are made of two or more atoms of different elements, such as water (H2O) and methane (CH4). Atoms are not drawn to scale. Molecules of compounds have atoms of two or more different elements.

7 0

2 years ago

7. Write the equation for the positron emission of barium-127.

ziro4ka [17]

The reaction is given by

$\\ \rm\Rrightarrow {}^{127}_{56}Ba\longrightarrow {}^{0}_{+1}\beta+{}^{127}_{55}Cs$

Barium goes underneath beta decay to form Ceaseum

Cs is very mellable element
It can melt on your hand

8 0

2 years ago

A sample of 211 g of iron (III) bromide is reacted with

Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

$\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714$

Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

$\tt n=\dfrac{186}{78.0452}=2.38$

Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :

$\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793$

So FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

$\tt \dfrac{6}{3}\times 0.714=1.428$

6 0

3 years ago

Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.

madreJ [45]

Answer:

$Molar \ solubility=3.12x10^{-5}M$

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

$CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-$

The equilibrium expression is:

$Ksp=[Ca^{2+}][F^-]^2$

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent $x$ is computed as follows:

$3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M$

Thus, the molar solubility equals the reaction extent $x$ , therefore:

$Molar \ solubility=3.12x10^{-5}M$

Regards.

4 0

3 years ago