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rodikova [14]
3 years ago
15

Question 5(Multiple Choice Worth 3 points)

Chemistry
1 answer:
Katarina [22]3 years ago
4 0

Answer:

\sqrt{ | {o2}^{2} | }  \times \frac{?}{?}  \sqrt[?]{?}  \times \frac{?}{?}  \sqrt[ |?| ]{?}  \sqrt{?}  \times  \frac{?}{?}

Explanation:

exactly

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How many significant digits does the number 700 have?
OLga [1]

Answer: 700 has one significant figure which is 7.

Explanation: These are some rules for significant figures

•All non-zero digits are significant: 1,2,3,4,5,6,7,8,9

•Zero between non-zero digits are significant: They are three significant figures in 203.

•Leading zeros are not significant: There are two significant figures in 0.56.

•Trailing zero to the right of decimal are significant. There are four significant figures in 62.00

•Trailing zeros in a whole number with the decimal shown are significant: This makes "700." three significant figures.

•Trailing zeros in a whole number with no decimal shown are not significant: This makes 700 one significant figure.

5 0
3 years ago
Read 2 more answers
What way can you conserve water?
Svetradugi [14.3K]
Run clothes washer only when its full
7 0
3 years ago
A student uses 0. 0821 LĂ˘â‚¬Ë atm/mol Ă˘â‚¬Ë K as the value of the gas constant. What is most likely true about the variables i
xz_007 [3.2K]

The variables in the ideal gas constant has V as the unit of liters and T has the unit of Kelvin. Thus, option C is correct.

The gas constant in an ideal gas equation has been the value of the energy absorbed by 1 mole of an ideal gas at standard temperature and pressure.

The value of R has been dependent on the units of volume, temperature and pressure of the ideal gas.

The given value of R has been 0.0821 L.atm/mol.K

The unit in gas constant has been L (Liter) for volume (V).

The unit of pressure (P) has been atm.

The unit of temperature (T) has been Kelvin  (K).

Thus the gas law constant used by student has V has the unit of liters and T has the unit of Kelvin. Thus, option C is correct.

For more information about the gas constant, refer to the link:

brainly.com/question/24814070

6 0
3 years ago
A piston in a heat engine does 500 joules of work, and 1400 joules of heat are added to the system. Determine the change in inte
san4es73 [151]

Answer : The change in internal energy is, 900 Joules.

Solution : Given,

Heat given to the system = +1400 J

Work done by the system = -500 J

Change in internal energy is equal to the sum of heat energy and work done.

Formula used :

\Delta U=q+w

where,

\Delta U = change in internal energy

q = heat energy

w = work done

As per question, heat is added to the system that means, q is positive and work done by the system that means, w is negative.

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

\Delta U=(+1400J)+(-500J)=900J

Therefore, the change in internal energy is 900 J.

The change in internal energy depends on the heat energy and work done. As we will change in the heat energy and work done, then changes  will occur in the internal energy. Hence, the energy is conserved.

8 0
3 years ago
Use the balanced equation from number four above to determine the limiting reactant if we had a mixture of 5.0 moles of Fe and 4
Tom [10]

Answer:

2Fe + 3O₂  →  2Fe₂O₃

Limiting reactant: O₂

2Fe + O₂ → 2FeO

Limiting reactant: Fe

Explanation:

2Fe + 3O₂  →  2Fe₂O₃

2Fe + O₂ → 2FeO

These are the possible reactions:

In first case, 2 moles of Fe need 3 mol of oyxgen to react

If I have 5 moles of Fe, I will need (5 .3)/2 = 7.5 moles of O₂

Then, the oxygen is my limiting ( I only have 4 moles)

3 moles of O₂ need 2 moles of Fe

If I have 4 moles of O₂, I will need (4 .2)/3 = 2.66 moles (I have 5)

Fe, is the reactant in excess.

For second case, 2 moles of Fe need 1 mol of O₂, to react.

If I have 5 moles of Fe, I will need ( 5 .1) / 2 =2.5 moles of O₂

I have 4 moles of oxygen, so now it is my excess.

1 mol of O₂ need 2 moles of Fe, to react

If I have 4 moles of O₂, I will need the double of amount, 8 moles of Fe.

I have 5 moles, then the Fe is my limtiing reactant.

8 0
3 years ago
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