Who ever gets right gets brainliest

Sanjay is learning about the elements that make up Earth’s crust. Ms. Richards, his Earth science teacher, gives him an element

oksano4ka [1.4K]

Answer:

Iron

Hope this helps! Stay safe!

(Plz vote me as brainliest)

3 0

3 years ago

What is the force on a 1000 kg elevator that is falling freely at 9.8/sec2

Mamont248 [21]

1,000 x 9.8 = 9800 Newtons

hope this helps

5 0

3 years ago

According to the image, identify the number of neutrons in the most common isotope of aluminum.

wolverine [178]

Answer:

The number of neutron in the Aluminium Isotope is :

B. 14

Explanation:

Isotopes : These are the atoms which have same atomic number but have different mass number.

<u>This image shows the average atomic mass of Al element because it is in decimals</u>.

Atomic mass = 26.98154

(Note : mass number of single isotope can never be in decimals)

It is the average of mass of different isotopes of Al

Major Isotopes of $_{13}^{26.98154}\textrm{Al}$ are :

$_{13}^{26}\textrm{Al}$ ......atomic mass = 26
$_{13}^{27}\textrm{Al}$ .......atomic mass = 27

mass of Al given in image(26.98) is nearly equal to mass of 2nd isotope(27)

mass of $_{13}^{26.98154}\textrm{Al}\ \approx 27$

Now calculate the neutron in $_{13}^{27}\textrm{Al}$

Number of neutron = mass number - atomic number

= 27 - 13

Number of neutron = 14

(Atomic mass is same as mass number)

5 0

3 years ago

Number 2 & 3 please !!<3

Goshia [24]

Answer:just for the ponits

Explanation:ima be a jenna

8 0

2 years ago

The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these com

Elanso [62]

Answer:

(a) $Ksp=4.50x10^{-7}$

(b) $Ksp=1.55x10^{-6}$

(c) $Ksp=2.27x10^{-12}$

(d) $Ksp=1.05x10^{-22}$

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) $BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)$

$Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}$

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

$Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}$

(B) $Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)$

$Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}$

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

$Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}$

(C) $NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)$

$Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}$

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

$Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}$

(D) $La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)$

$Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}$

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

$Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}$

Best regards.

7 0

3 years ago