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3 years ago

The time it takes to run a kilometer depends on the amount of exercise a person gets.

1 answer:

3 0

<span>The amount of exercise that a person gets would be the independent variable, as it would impact the amount of time it would take to run the kilometer, with time being the dependent variable.</span>

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Substances that cannot be separated and found on a periodic table

GaryK [48]

Substances that cannot be separated and found on a periodic table are elements.

4 0

3 years ago

Read 2 more answers

Formed when an amine is combined with a carboxyl group

alekssr [168]

Hello.

The answer is C.Amine

When an amine is combined (reacted) with a carboxyl group, an AMIDE + water is formed, and if you carry on heating under a vacuum, an imidazoline is formed.

Have a nice day

7 0

3 years ago

A chemical reaction produced 10.1 cm3 of nitrogen gas at 23 °C and 746 mmHg. What is the volume of this gas if the temperature a

adell [148]

Answer:

volume of gas = 9.1436cm³

Explanation:

We will only temperature from °C to K since the conversion is done by the addition of 273 to the Celsius value.

Its not necessary to convert pressure and volume as their conversions are done by multiplication and upon division using the combined gas equation, the factors used in their conversions will cancel out.

V1 =10.1cm³ , P1 =746mmHg, T1=23°C =23+273=296k

V2 =? , P2 =760mmmHg , T2=0°C = 0+273 =273K

Using the combined gas equation to calculate for V2;

$\frac{V1P1}{T1}=\frac{V2P2}{T2} \\ re-arranging, \\V2 =\frac{V1P1T2}{P2T1}$

$V2 =\frac{10.1*746*273}{760*296}$

V2=9.1436cm³

6 0

3 years ago

Help please.

Alexxx [7]

Answer:

this is my old account and these still dont have answers

Explanation:

8 0

2 years ago

CAN SOMEONE HELP ME PLZ AND THANKS WILL MARK U AS BRAINLIEST

jekas [21]

Explanation:

2. $2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$

First, we need to find the number of moles of $CO_2$ at 300K and 1.5 atm using the ideal gas law:

$n= \dfrac{PV}{RT}= \dfrac{(1.5\:\text {atm})(33\:L)}{(0.082\:\text{L-atm/mol-K})(300K)}$

$=2.0\:\text{mol}\:CO_2$

Now use the molar ratios to find the number of moles of ethane to produce this much $CO_2$ .

$2.0\:\text{mol}\:CO_2 \times \left(\dfrac{2\:\text{mol}\:C_2H_6}{4\:\text{mol}\:CO_2}\right)$

$=1.0\:\text{mol}\:C_2H_6$

Finally, convert this amount to grams using its molar mass:

$1.0\:\text {mol}\:C_2H_6 \times \left(\dfrac{30.07\:\text g\:C_2H_6}{1\:\text{mol}\:C_2H_6} \right)$

$=30.1\:g\:C_2H_6$

3. $3Zn + 2H_3PO_4 \rightarrow 3H_2 + Zn_3(PO_4)_2$

Convert 75 g Zn into moles:

$75\:\text g\:Zn \times \left(\dfrac{65.38\:\text g\:Zn}{1\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:Zn$

Then use the molar ratios to find the amount of H2 produced.

$1.1\:\text{mol}\:Zn \times \left(\dfrac{3\:\text{mol}\:H_2}{3\:\text{mol}\:Zn}\right)=1.1\:\text{mol}\:H_2$

Now use the ideal gas law $PV=nRT$ to find the volume of H2 produced at 23°C and 4 atm:

$V= \dfrac{nRT}{P}= \dfrac{(1.1\:\text{mol}\:H_2)(0.082\:\text{L-atm/mol-K})(296K)}{4\:\text{atm}}$

$=8.9\:\text L\:H_2$

8 0

2 years ago