Can someone help me with this?

3. What is the density of a 100 grams (g) box that displaces 20 mL of water?

Zolol [24]

Answer: the density is 997 kg

Explanation:

8 0

3 years ago

What makes a heterogeneous mixture different from a homogeneous mixture? The components are mixed unevenly instead of evenly wit

bazaltina [42]

<span>is a mixture that composes of components that aren't uniform or they have localized regions that all have different properties. Despite the term appearing to be highly scientific, there are various common substances that are heterogeneous mixtures

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8 0

3 years ago

What are the subatomic particles of barium

ivolga24 [154]

Electrons (negative charge, and orbits the nucleus), Protons (positive charge, and is in the nucleus) and Neutrons (no charge, and is in the nucleus)

7 0

4 years ago

Which one of the following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 → 4NO2 + 6H2O Which one of t

Veseljchak [2.6K]

Answer : All of the above are valid expressions of the reaction rate.

Explanation :

The given rate of reaction is,

$4NH_3+7O_2\rightarrow 4NO_2+6H_2O$

The expression for rate of reaction for the reactant :

$\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}$

$\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}$

The expression for rate of reaction for the product :

$\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}$

From this we conclude that, all the options are correct.

3 0

3 years ago

If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?

ElenaW [278]

Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$ where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation: $T_C+273=T_K$

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, $T_1 = 273[K]$ , and $T_2 = (49.4+273)[K]=322.4[K]$ .

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}$

$\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})$

$1810.04571428[mL]=V_2$

Adjusting for significant figures, this gives $V_2=1810[mL]$

4 0

2 years ago