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2 years ago

Metallic properties tend to increase in which direction on the periodic table

Chemistry

1 answer:

tresset_1 [31]2 years ago

3 0

Metallic properties head to the left.

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A mineral that helps in clotting of blood________.

sashaice [31]

Answer:

Vitamin K

Explanation:

this is the answer

7 0

3 years ago

Read 2 more answers

Atmospheric pressure at sea level is equal to a column of 760 mm Hg. Oxygen makes up 21 percent of the atmosphere by volume. The

Neporo4naja [7]

<u>Answer:</u> The partial pressure of oxygen is 160 mmHg

<u>Explanation:</u>

We are given:

Percent of oxygen in air = 21 %

Mole fraction of oxygen in air = $\frac{21}{100}=0.21$

To calculate the partial pressure of oxygen, we use the equation given by Raoult's law, which is:

$p_{O_2}=p_T\times \chi_{O_2}$

where,

$p_{O_2}$ = partial pressure of oxygen = ?

$p_T$ = total pressure of air = 760 mmHg

$\chi_{O_2}$ = mole fraction of oxygen = 0.21

Putting values in above equation, we get:

$p_{O_2}=760mmHg\times 0.21\\\\p_{O_2}=160mmHg$

Hence, the partial pressure of oxygen is 160 mmHg

8 0

2 years ago

How much CaCO3 would have to be decomposed to produce 247 g of CaO

vovangra [49]

441 g CaCO₃ would have to be decomposed to produce 247 g of CaO

<h3>Further explanation</h3>

Reaction

Decomposition of CaCO₃

CaCO₃ ⇒ CaO + CO₂

mass CaO = 247 g

mol of CaO(MW=56 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{247}{56}\\\\mol=4.41$

From equation, mol ratio CaCO₃ : CaO = 1 : 1, so mol CaO :

$\tt \dfrac{1}{1}\times 4.41=4.41$

mass CaCO₃(MW=100 g/mol) :

$\tt mass=mol\times MW\\\\mass=4.41\times 100\\\\mass=441~g$

6 0

2 years ago

This fine-grained dark colored igneous rock is most likely to

Serggg [28]

Most likely basalt....

6 0

2 years ago

Consider the second-order reaction:

kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)
$

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

5 0

3 years ago