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3 years ago

11

Why are fluxes added to sand

1 answer:

adell [148]3 years ago

5 0

Answer:

no idea hahahahahhaaha

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7. Which of these is not matter?<br> a. a cloud<br> c. sunshine<br> b. your hair<br> d. the sun

PilotLPTM [1.2K]

Hair ????????????????!?????

3 0

3 years ago

When the volume of a gas is

Montano1993 [528]

Answer:

$\boxed {\boxed {\sf 232.9 \textdegree C}}$

Explanation:

This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

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

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

$\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}$

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

$\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}$

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

$250 \ mL * T_2 = 137 \textdegree C * 425 \ mL$

Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.

$\frac{250 \ mL * T_2}{250 \ mL}=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}$

$T_2=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}$

The units of milliliters (mL) cancel.

$T_2=\frac{ 137 \textdegree C * 425 }{250 }$

$T_2= \frac{58225}{250} \textdegree C$

$T_2=232.9 \textdegree C$

The temperature changes to <u>232.9 degrees Celsius.</u>

3 0

3 years ago

Write the overall reaction for cellular respiration

goblinko [34]

C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O

7 0

3 years ago

An inventor claims to have devised a closed cyclic engine which exchanges heat with cold and hot reservoirs at 25 and 350 ◦C, re

kati45 [8]

Explanation:

The given data is as follows.

$T_{c} = 25^{o}C$ , $T_{h} = 350^{o}C$

Work produced per kJ of heat extracted from hot reservoir = 0.45 kJ = Efficiency

If the device is Carnot cycle then its efficiency will be maximum and its value will be equal to $[1 - (\frac{T_{c}}{T_{h}} )]$

Using this relation we will calculate the efficiency as follows.

Efficiency = $[1 - (\frac{T_{c}}{T_{h}} )]$

= $1 - (\frac{25}{350})$

= 0.928

Hence, it means that this type of device is possible and the claim is also believable.

7 0

3 years ago

In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2

valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

$\dfrac{80}{17}=4.706$

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

$\tt \dfrac{120}{32}=3.75$

Mol ratio of reactants(to find limiting reatants) :

$\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)$

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

$\tt \dfrac{6}{5}\times 3.75=4.5$

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

$\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%$

6 0

3 years ago