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Fofino [41]
3 years ago
12

The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's moti

on?
OA. It's not moving
OB. It's moving at a constant speed.
OC. It's moving at a constant velocity.
OD
It's speeding up.
Chemistry
1 answer:
marishachu [46]3 years ago
8 0
B because a horizontal line goes straight across which shows a constant rate
Hope this helps!
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How much heat is required to vaporize 25g of water at 100*c
ElenaW [278]

Answer:

Heat required = 13,325 calories or 55.75 KJ.

Explanation:

To convert a water to steam at 100 degree celsius to vapor, we have to give latent heat of vaporization to water

Which equals ,

Q = mL,

Where, m is the mass of water present

           L = specific latent heat of vaporization

Here , m= 25 gram

L equals to 533 calories (or 2230 Joules)

So, Q = 25×533 = 13,325 Calories

Or , Q = 55,750 Joules = 55.75 KJ

so, Heat required = 13,325 calories or 55.75 KJ.

4 0
3 years ago
What is the mass of 1.45 moles of silver sulfate?
tatuchka [14]

Answer:

449.5 g

Explanation:

Silver sulfate- Ag2SO4

M(Ag)=107 g/mol => M(Ag2)=214 g/mol

M(S)=32 g/mol

M(O)=16 g/mol => M(O4)=64 g/mol

M(Ag2SO4)=310 g/mol

n=1.45 mol

m(Ag2SO4)=M(Ag2SO4)*n=310 g/mol *1.45 mol= 449.5 g

3 0
3 years ago
Read 2 more answers
Eading and
4vir4ik [10]

Answer: 25.8 g of Cl_2 will be produced from the decomposition of 73.4 g of AuCl_3

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} AuCl_3=\frac{73.4g}{303g/mol}=0.242moles

The balanced chemical reaction is:

2AuCl_3\rightarrow 2Au+3Cl_2  

According to stoichiometry :

2 moles of AuCl_3 produce =  3 moles of Cl_2

Thus 0.242 moles of  will produce= \frac{3}{2}\times 0.242=0.363mol of Cl_2

Mass of Cl_2= moles\times {\text {Molar mass}}=0.363mol\times 71g/mol=25.8g

Thus 25.8 g of Cl_2 will be produced from the decomposition of 73.4 g of AuCl_3

5 0
2 years ago
Help me with this this is something i dont know and its not on here please help meeeeeee
denis-greek [22]

Answer:

it would be option A

Explanation:

This is becuase if you look at the chart you can see tyhat the group of rats that got feed to vitamans did gain more wati then the ones on the normal diet.

8 0
2 years ago
The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio
iren2701 [21]

Answer : The correct option is, (B) CO_2

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of unknown gas = 11.9\text{ mL }min^{-1}

R_2 = rate of effusion of oxygen gas = 14.0\text{ mL }min^{-1}

M_1 = molar mass of unknown gas  = ?

M_2 = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}

M_1=44.2g/mole

The unknown gas could be carbon dioxide (CO_2) that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide (CO_2)

5 0
3 years ago
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