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Aleksandr-060686 [28]

3 years ago

8

A mole of anything is how many of that thing?

1 answer:

ikadub [295]3 years ago

7 0

1 mole is equal to the atomic mass of the element/compound in grams

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What four things affect the tides

S_A_V [24]

I know three
The Moon
The Wind
and Earths Rotation.

Gravity Maybe one thing but i am not sure

4 0

3 years ago

Assume a gasoline is isooctane, which has a density of 0.692 g/ml. What is the mass of 3.8 gal of the gasoline (1 gal = 3.78 l)?

sveticcg [70]

Density is the ratio of mass to the volume.

The mathematical expression is given as:

$density=\frac{mass}{volume}$

Now, density of isooctane = $0.692 g/mL$

Volume = $3.8 gal$

Since, 1 gallon = 3.78 L

So, 3.8 gal = $3.78 L\times 3.8$

= $14.364 L$

As, 1 L = 1000 mL

Therefore, $14.364 L$ = $14.364 L\times 1000 mL$

Volume in mL = $14364 mL$

Put the values,

$0.692 g/mL=\frac{mass}{14364 mL}$

$m = 0.692 g/mL\times 14364 mL$

= $9939.888 g$

Hence, mass of 3.8 gal of the gasoline is $9939.888 g$ .

6 0

3 years ago

If mixture didn't exist how would life be different<br><br>PLEASEEE QUICKKKK

Nataliya [291]

Answer:

boring

Explanation:

without meaning

8 0

3 years ago

Convert all of the following lengths to millimeters.

xenn [34]

A) 120 mm
B) 127 mm
C) 914.4 mm
D) 1000 mm
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8 0

2 years ago

Read 2 more answers

A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin

lorasvet [3.4K]

Answer : The mass of the water in two significant figures is, $3.0\times 10^1g$

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron metal = $0.45J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of iron metal = 32.3 g

$m_2$ = mass of water = ?

$T_f$ = final temperature of mixture = $59.2^oC$

$T_1$ = initial temperature of iron metal = $21.9^oC$

$T_2$ = initial temperature of water = $63.5^oC$

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

$32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC$

$m_2=30.16g\approx 3.0\times 10^1g$

Therefore, the mass of the water in two significant figures is, $3.0\times 10^1g$

3 0

3 years ago