, 24.01.2020emanueln

This is not a question that can be answered

Explanation:

Explanation:

Just trying to get some points

Explanation:

i have the same problem

B

Explanation:

a) Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

Explanation:

a) Statement A : 2.567km to two significant figures.

2.567km 2. S.F = 2.6km

Statement B : 2.567km to three significant figures.

2.567km 3 S.F = 2.57km

Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) statement A: (2.567 km + 3.146km) to 2 S.F

(2.567km + 3.146km) = 5.713km to 2 S.F = 5.7km

Statement B : (2.567 km, to two significant figures) + (3.146 km, to two significant figures).

2.567km to 2 S.F = 2.6km

3.146km to 2 S.F = 3.1km

2.6km + 3.1km = 5.7km

Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

a) rounded to the ones place and 2 significant figures.

b) rounded to tenths place and 1 significant figure.

Step-by-step explanation:

Significant figures are the number of needed place values for accuracy. Round each number according to the least number of significant figures within the problem.

A. 9 cm+2.8 cm = 11.8 cm rounds to 12 which has 2 significant figures. Round to the ones place value since 9 has the least number of significant figures.

B. 0.135 atm+0.6 atm = 0.735 atm rounds to 0.7 atm which has 1 significant figure. Zero preceding the value is not considered significant. Round to the tenths place value since 0.6 has the least number of significant figures.

a) 7.2 × 10³ J

b) -5.1 × 10² J

c) 7.7 × 10³ J

Explanation:

a)Use the wattage of the light bulb and the time it is on to calculate ΔU in joules (assume that the cylinder and light bulb assembly is the system and assume two significant figures). Express your answer using two significant figures.

The light bulb has a power of 100 W (100 J/s) and works during a time of 2.0 × 10⁻² hours. The change in the internal energy (ΔU) is:

We can calculate the work (w) using the following expression.

w = -P . ΔV

where,

P is the external pressure

ΔV is the change in the volume

We can calculate the heat (q) using the following expression.

ΔU = q + w

q = ΔU - w = 7.2 × 10³ J - (-5.1 × 10² J) = 7.7 × 10³ J

5 significant figures: 670.90 in.

4 significant figures: 8.314 J/mol-K, 6.022x10 23 mol-1, 0.6258 mg, 2.205 lb and 3270 ft.

3 significant figures: 40.7 g.

2 significant figures: 12 mL, 1.2 years, 7.0 m2, 50, cm and 90 m.

1 significant figure: 0,005 L.

Explanation:

Hello,

In this case, we proceed as follows:

* 8.314 J/mol-K has 4 significant figures since 8, 3, 1 and 4 are significant (nonzero).

* 6.022x10 23 mol-1 has 4 significant figures since 6, 0, 2 and 2 are significant including the zero as it is at the right of the first nonzero digit (6).

* 12 mL has 2 significant figures since 1 and 2 are significant (nonzero).

* 1.2 years has 2 significant figures since 1 and 2 are significant (nonzero).

* 7.0 m2 has 2 significant figures since 1 and 2 are significant (nonzero).

* 50, cm has 2 significant figures since 1 and 2 are significant (nonzero).

* 0,005 L has 1 significant figure since 5 is the only nonzero digit.

* 40.7 g has 3 significant figures since 4, 0 and 7 are significant including the zero as it is at the right of the first nonzero digit (4).

* 670.90 in has 5 significant figures since 6, 7, 0, 9 and 0 are significant including the zeros as they are at the right of the first nonzero digit (6).

* 0.6258 mg has 4 significant figures since the first zero is at the left of the first nonzero digit, therefore, it is not included.

* 2.205 lb has 4 significant figures since 2, 2, 0 and 5 are significant including the zero as it is at the right of the first nonzero digit (2).

* 3270 ft has 4 significant figures since 3, 2, 7 and 0 are significant including the zero as it is at the right of the first nonzero digit (3).

* 90 m has 2 significant figures since 1 and 2 are significant (nonzero).

Regards.

156 000: 3 significant figures

Explanation:

The rules to identify the correct number of significant figures in a number are:

- All the non-zero digits are always significant

- Final (trailing) zeros are only significant if they are after the decimal point

- All zeros between non-zero digits are significant

Now we can apply these rules to the given numbers:

156 000: 3 significant figures --> TRUE. Only 1, 5 and 6 are significant figures here.

0.00302: 2 significant figures --> FALSE. Here the significant figures are only 3, 0 and 2, so there are 3 significant figures

0.04300: 5 significant figures --> FALSE. Here the significant figures are 4, 3, 0, 0: so, 4 sign. figures.

3.0650: 4 significant figures --> FALSE. Here there are 5 significant figures: 3,0,6,5,0

1.04: 2 significant figures --> FALSE. Here there are 3 significant figures: 1,0,4

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f)  12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only  the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) =  4.886

(d) mol HA reacted = 150 x 10⁻³ L  x  x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) =  5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH  = √(kb x (A⁻)  where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA,  200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒   Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

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