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snow_tiger [21]

3 years ago

6

Gluteus maximus Hint: You're probably sitting on it. Need help ?!!

2 answers:

Rudiy273 years ago

8 0

Answer:

i honestly have no idea

Explanation:

charle [14.2K]3 years ago

8 0

Answer:

It's a muscle component near your thighs and hips everyone.

Explanation:

I don't really know what your question is, but uh...

The Hallux is another word for your big toe.

<em>Hope </em><em>it </em><em>helps!</em>

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Which step of the scientific method involves obtaining resources for your question?

djyliett [7]

I would say it is constructing a hypothesis, that is because a hypothesis is basically gathering information or resources to support what your question was. so i say it is B.

3 0

3 years ago

Read 2 more answers

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq)HCl(aq), as descr

ra1l [238]

Answer: 0.887 g of $MnO_2$ should be added to excess HCl(aq).

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 805 torr = 1.06 atm (760torr=1atm)

V = Volume of gas = 235 ml = 0.235 L

n = number of moles = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $25^0C=(25+273)K=298K$

$n=\frac{PV}{RT}$

$n=\frac{1.06atm\times 0.235L}{0.0820 L atm/K mol\times 298K}=0.0102moles$

$MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)$

According to stoichiometry:

1 mole of chlorine is produced by = 1 mole of $MnO_2$

Thus 0.0102 moles of chlorine is produced by = $\frac{1}{1}\times 0.0102=0.0102$ moles of $MnO_2$

Mass of $MnO_2$ = $moles\times {\text {Molar mass}}=0.0102mol\times 87g/mol=0.887g$

0.887 g of $MnO_2$ should be added to excess HCl(aq).

7 0

3 years ago

Suppose that some FeSCN2+ is added to the above solution to shift the equilibrium. When equilibrium is re-established, the follo

kiruha [24]

Explanation:

The given reaction equation will be as follows.

$[FeSCN^{2+}] \rightleftharpoons [Fe^{3+}] + [SCN^{-}]$

Let is assume that at equilibrium the concentrations of given species are as follows.

$[Fe^{3+}] = 8.17 \times 10^{-3}$ M

$[SCN^{-}] = 8.60 \times 10^{-3}$ M

$[FeSCN^{2+}] = 6.25 \times 10^{-2}$ M

Now, first calculate the value of $K_{eq}$ as follows.

$K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}$

= $\frac{8.17 \times 10^{-3} \times 8.60 \times 10^{-3}}{6.25 \times 10^{-2}}$

= $11.24 \times 10^{-4}$

Now, according to the concentration values at the re-established equilibrium the value for $[FeSCN^{2+}]$ will be calculated as follows.

$K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}$

$11.24 \times 10^{-4} = \frac{8.12 \times 10^{-3} \times 7.84 \times 10^{-3}}{[FeSCN^{2+}]}$

$[FeSCN^{2+}] = 5.66 \times 10^{-2}$ M

Thus, we can conclude that the concentration of $[FeSCN^{2+}]$ in the new equilibrium mixture is $5.66 \times 10^{-2}$ M.

7 0

3 years ago

An atom has 24 protons, 22 electrons, and 28 neutrons. What is the ATOMIC # of the atom?

Over [174]

22 would be the answer

7 0

3 years ago

2503(g) → 2502(g) O2(g)

Verizon [17]

Answer:

7.146

Explanation:

use the equilibrium equation

6 0

3 years ago