I

I have attached all the problems, but really if you just do one so I understand how to do it, that would be great!

Ivahew [28]

Answer:

Answer of question a is 345J.

Explanation:

In question a following is given in data:

-mass of iron (m) = 10.0 g

-temperature (ΔT) = final temperature- initial temperature= 100-25= 75 degree Celsius

-Specific Heat capacity of iron (c)= 0.46J/g°C.

Heat (Q)=?

Solution:

Formula for Heat is :

Q=m x c x ΔT

Q= 10 x 0.46 x 75

Q= 345 J.

so, 345 joules of heat is needed to increase the temperature of 10 grams of iron.

From the above formula all other questions can easily be solved from the same procedure.

4 0

3 years ago

Which term describes the molecule shown below?

Aleks04 [339]

Answer:

Saturated alkyne

Explanation:

6 0

3 years ago

N2H4 + N2O4 --> N2 + H2O

ahrayia [7]

Answer:

The limiting reagent is N2O4
14,09g

Explanation:

First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄ $6N_{2} + 4H_{2}O$

Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

$molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

$molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

$molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$

$molH_{2} O = 0,783 mol$

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴ $MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}$

$MassOfH_{2}O = 14,09g$

5 0

3 years ago

A sample of H2 gas (12.28 g) occupies 100.0 L at 400.0 K and 2.00 atm. A sample weighing 9.49 g occupies ________ L at 353 K and

lutik1710 [3]

Considering the ideal gas law, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P× V = n× R× T

In this case, you know:

P= 2 atm
V= ?
n= $9.49 gramsx\frac{1 mole}{2 grams} = 4.745moles$ being 2g/mole the molar mass of H2, that is, the amount of mass that a substance contains in one mole.
R= 0.082 $\frac{atmL}{molK}$
T= 353 K

Replacing:

2 atm× V = 4.745 moles× 0.082 $\frac{atmL}{molK}$ × 353 K

Solving:

V = (4.745 moles× 0.082 $\frac{atmL}{molK}$ × 353 K)÷ 2 atm

Finally, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.

Learn more:

brainly.com/question/4147359?referrer=searchResults

5 0

2 years ago

Which is a negative result of eating too many lipids plz help

fenix001 [56]

Eating too much lipids is hazardous to the health. It has a lot of negative effects. One is cardiovascular complications, too much of lipids can cause high blood cholesterol, arterial hardening, heart diseases and stroke. Also, obesity is possible. Eating too much lipids is associated with metabolism abnormalities.

5 0

3 years ago

Read 2 more answers