Ask question

Ask question

All categories

3 years ago

9

Sarah cycles 30 mi to her grandmother's house at a steady speed of 10 mph. If she leaves home at 2:00 PM what time will she arri

ve?

1 answer:

gulaghasi [49]3 years ago

4 0

5:00 pm will be the time she arrives

You might be interested in

The sample of 15.0 g of KCl is dissolved into a solution with a total volume of 250.0 mL. What is the molarity of KCl in the sol

lina2011 [118]

Answer:

0.805 M.

Explanation:

Hello!

In this case, since the molarity of a solution is computing by dividing the moles of solute over the volume of solution in liters (M=n/V), for 15.0 g of potassium chloride (74.55 g/mol) we compute the corresponding moles:

$n=15.0gKCl*\frac{1molKCl}{74.55gKCl}=0.201molKCl$

Next, since the volume is 0.2500 in liters, the molarity turns out:

$M=\frac{0.201mol}{0.2500L} \\\\M=0.805M$

Best regards!

3 0

2 years ago

Can someone tell me how to solve this step by step? I got the answer but I want to know how to solve it I have a test !!!!

UNO [17]

C becuase that is the one that you would have to do becuase that is the only option

6 0

2 years ago

Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2

PIT_PIT [208]

<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ$

Hence, the standard heat for the given reaction is -138.82 kJ

3 0

2 years ago

PLEASE HELP MEEE!!!!

Assoli18 [71]

It would be 35.8 Calories or calories. Not sure about that part. Hope this helps though.

3 0

2 years ago

Determine how many ml of water you need to remove, by evaporation, if you have a 500 ml of 10.20 M HNO3 dilute solution and you

Ludmilka [50]

The total volume of water that would be removed will be 75 mL

<h3>Dilution equation</h3>

Using the dilution equation:

M1V1 = M2V2

In this case, M1 = 500 mL, V1 = 10.20 M, M2 = 12 M

Substitute:

V2 = 500 x 10.20/12

= 425 mL

The final volume in order to arrive at 12 M HNO3 would be 425 mL from the initial 500 mL. Thus, the total amount of water that will be removed by evaporation can be calculated as:

500 - 425 = 75 mL

More on dilution can be found here: brainly.com/question/7208939

7 0

2 years ago