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3 years ago

5

Compare and contrast how wind glaciers abrade rock?

1 answer:

pochemuha3 years ago

3 0

Wind abrades rock by sandblasting, this is the process in which wind causes the

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Which of the following elements is a metal in the periodic table

OverLord2011 [107]

Answer:

12 My Magnesium

13 Al Aluminum

19 K Potassium

20 Ca Calcium

Explanation:

I think

8 0

3 years ago

If it requires 23.4 milliliters of 0.65 molar barium hydroxide to neutralize 42.5 milliliters of nitric acid, solve for the mola

rodikova [14]

Volume Ba(OH)2 = 23.4 mL in liters :

23.4 / 1000 => 0.0234 L

Molarity Ba(OH)2 = 0.65 M

Volume HNO3 = 42.5 mL in liters:

42.5 / 1000 => 0.0425 L

number of moles Ba(OH)2 :

n = M x V

n = 0.65 x 0.0234

n = 0.01521 moles of Ba(OH)2

Mole ratio :

<span>Ba(OH)2 + 2 HNO3 = Ba(NO3)2 + 2 H2O
</span>
1 mole Ba(OH)2 ---------------- 2 moles HNO3
0.01521 moles ----------------- moles HNO3

moles HNO3 = 0.01521 x 2 / 1

moles HNO3 = 0.03042 / 1

= 0.03042 moles HNO3

Therefore:

M ( HNO3 ) = n / volume ( HNO3 )

M ( HNO3 ) = 0.03042 / 0.0425

M ( HNO3 ) = 0.715 M

5 0

3 years ago

A pycnometer is a precisely weighted vessel that is used for highly accurate density determinations. Suppose that a pycnometer h

dexar [7]

Answer:

5.758 is the density of the metal ingot in grams per cubic centimeter.

Explanation:

1) Mass of pycnometer = M = 27.60 g

Mass of pycnometer with water ,m= 45.65 g

Density of water at 20 °C = d = $998.2 kg/m^3$

1 kg = 1000 g

$1 m^3=10^6 cm^3$

$998.2 kg/m^3=\frac{998.2 \times 1000 g}{10^6 cm^3}=0.9982 g/cm^3$

Mass of water ,m'= m - M = 45.65 g - 27.60 g =18.05 g

Volume of pycnometer = Volume of water present in it = V

$Density=\frac{Mass}{Volume}$

$V=\frac{m'}{d}=\frac{18.05 g}{0.9982 g/cm^3}=18.08 cm^3$

2) Mass of metal , water and pycnometer = 56.83 g

Mass of metal,M' = 9.5 g

Mass of water when metal and water are together ,m''= 56.83 g - M'- M

56.83 g - 9.5 g - 27.60 g = 19.7 g

Volume of water when metal and water are together = v

$v=\frac{m''}{d}=\frac{19.7 g}{0.9982 g/cm^3}=19.73 cm^3$

Density of metal = d'

Volume of metal = v' = $\frac{M'}{d'}$

Difference in volume will give volume of metal ingot.

v' = v - V

$v'=19.73 cm^3-18.08 cm^3=$

$v'=1.65 cm^3$

Since volume cannot be in negative .

Density of the metal =d'

= $d'=\frac{M'}{v'}=\frac{9.5 g}{1.65 cm^3}=5.758 g/cm^3$

5 0

3 years ago

What element has an electron<br> configuration of 2-8-8-1:

Afina-wow [57]

Ойлголоо, уучлаарай Ойлголоо, уучлаарай /; coo

4 0

3 years ago

The Ksp of CaF2 is 4.0 x 10-11. What is the solubility of CaF2 in 0.10 M Ca(NO3)2? Give your answer using scientific notation an

Ilya [14]

Answer:

$1.0\times 10^{-5} M$ is the solubility of $CaF_2$ in 0.10 M $Ca(NO_3)_2$ .

Explanation:

Concentration of calcium nitrate = $[Ca(NO_3)_2]=0.10 M$

$Ca(NO_3)_2\rightarrow Ca^{2+}+2NO_3^{-}$

$[Ca(NO_3)_2]=0.10 M=[Ca^{2+}]$

Solubility product of the calcium fluoride = $K_{sp}=4.0\times 10^{-11}$

Solubility of calcium fluoride in calcium nitrate solution :

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

(0.10+S) (2S)

The expression of solubility is given by :

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

$4.0\times 10^{-11}=(S+0.10)(2S)^2$

Solving for S:

$S=1.0\times 10^{-5} M$

$1.0\times 10^{-5} M$ is the solubility of $CaF_2$ in 0.10 M $Ca(NO_3)_2$ .

3 0

3 years ago