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3 years ago

If an object has a Mass of 11g and a volume fo 13mil, what is its density?

Chemistry

1 answer:

Nimfa-mama [501]3 years ago

8 0

Answer:

$\boxed {\boxed {\sf B. \ d \approx 0.85 \ g/mL}}$

Explanation:

Density is found by dividing the mass by the volume.

$d=\frac{m}{v}$

The mass of the object is 11 grams and the volume is 13 milliliters.

$m= 11 \ g \\v= 13 \ mL$

Substitute the values into the formula.

$d= \frac{11 \ g}{13 \ mL}$

Divide.

$d=0.846153846 \ g/mL$

Round to the nearest hundredth. The 6 in the thousandth place tells us to round the 4 to a 5.

$d \approx 0.85 \ g/mL$

The density is about <u>0.85 grams per milliliter.</u>

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The oxidation of ammonia produces nitrogen and water via the following reaction: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l) Suppose the

Sonbull [250]

Answer:

The rate of consumption of $NH_{3}$ is 2.0 mol/L.s

Explanation:

Applying law of mass action to this reaction-

$-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N_{2}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}$

where $-\frac{\Delta [NH_{3}]}{\Delta t}$ represents rate of consumption of $NH_{3}$ , $-\frac{\Delta [O_{2}]}{\Delta t}$ represents rate of consumption of $O_{2}$ , $\frac{\Delta [N_{2}]}{\Delta t}$ represents rate of formation of $N_{2}$ and $\frac{\Delta [H_{2}O]}{\Delta t}$ represents rate of formation of $H_{2}O$ .

Here rate of formation of $H_{2}O$ is 3.0 mol/(L.s)

From the above equation we can write-

$-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}$

Here $\frac{\Delta [H_{2}O]}{\Delta t}=3.0 mol/(L.s))$

So, $-\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\frac{\Delta [H_{2}O]}{\Delta t}$

Hence, $-\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\times 3.0 mol/(L.s)=2.0 mol/(L.s)$

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3 years ago

Which is most likely to provide the best data on volcanic activity over the last 2000 years?

dimaraw [331]

Answer:

By the time the Vesuvius eruption sputtered to an end the next day, Pompeii was buried under millions of tons of volcanic ash. About 2,000 Pompeiians were dead, but the eruption killed as many as 16,000 people overall. By

the time the Vesuvius eruption sputtered to an end the next day, Pompeii was buried under millions of tons of volcanic ash. About 2,000 Pompeiians were dead, but the eruption killed as many as 16,000 people overall.

Explanation:

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3 years ago

The net ionic equation for the reaction between aqueous sulfuric acid and aqueous sodium hydroxide is ________.

Kaylis [27]

Molecular equation:
$H_{2}SO_{4}(aq) + 2NaOH(aq)--\ \textgreater \ Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Complete ionic equation:

2H^{+}+SO_{4}^{2-} + 2Na^{+} +2OH^{-}--\ \textgreater \ 2Na^{+}+SO_{4}^{2-} + 2H_{2}O(l)


Near 2H^{+},SO_{4}^{2-} , 2Na^{+} , OH^{-} 

should be written (aq).

Remove ions that do not change from both sides

2H^{+} + 2OH^{-}=2H_{2}O

Net ionic equation:

H^{+}(aq) + OH^{-}(aq)=H_{2}O(l)

Answer is D.

$

8 0

3 years ago

A 40.2 g sample of a metal heated to 99.3Â°C is placed into a calorimeter containing 120 g of water at 21.8Â°C. The final temper

aliina [53]

Answer:

B) Iron (c=0.45 J/g°C)

Explanation:

Given that:-

Heat gain by water = Heat lost by metal

Thus,

$m_{water}\times C_{water}\times (T_f-T_i)=-m_{metal}\times C_{metal}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,

$m_{water}\times C_{water}\times (T_f-T_i)=m_{metal}\times C_{metal}\times (T_i-T_f)$

For water:

Mass = 120 g

Initial temperature = 21.8 °C

Final temperature = 24.5 °C

Specific heat of water = 4.184 J/g°C

For metal:

Mass = 40.2 g

Initial temperature = 99.3 °C

Final temperature = 24.5 °C

Specific heat of metal = ?

So,

$120\times 4.184\times (24.5-21.8)=40.2\times C_{metal}\times (99.3-24.5)$

$40.2C_{metal}\left(99.3-24.5\right)=120\times \:2.7\times \:4.184$

$40.2C_{metal}\left(99.3-24.5\right)=1355.616$

$C_{metal}=0.45\ J/g^0C$

<u>This value corresponds to iron. Thus answer is B.</u>

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3 years ago

What are the elements in group 3-12 called

Wittaler [7]

Answer:

Transition Metals

Explanation:

The elements in groups 3-12 are called Transition Metals. These groups contain metals that usually form multiple cations. All other groups on the table (1, 2, 13-18) are called Main Group Elements.

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3 years ago