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3 years ago

How does gravity help to demonstrate a cause-and-effect relationship?

Chemistry

2 answers:

Zina [86]3 years ago

7 0

The cause of no gravity means the effect will be we float.

bazaltina [42]3 years ago

6 0

Answer:

Gravity helps show a cause and effect by demostarting how one object will have an effect on another object.

Explanation:

If you bounce a ball, it will fly back up, but it will fall down again because of gravity. The cause is GRAVITY and the effect is the ball FALLINg. Now, because o fgravity, when the ball bounces up again, it will only bounce 1/2 the hieght, having another cause and effect. As gravity pushes down on objects, the more an object is effected by it, the more presure it will have.

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A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago

Sulfur has an atomic mass of 32 u and Avogadro's number is 6 × 1023 formula units/mole. The number of sulfur atoms in 1 g of sul

Vikki [24]

Answer:

The number of sulfur atoms in 1 g of sulfur is:- $1.875\times 10^{22}$ atoms

Explanation:

Avogadro’s number represent the number of the constituent particles which are present in one mole of the substance. It is named after scientist Amedeo Avogadro and is denoted by $N_0$ .