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3 years ago

8

High self-monitors prefer situations in which clear expectations exist regarding how they're supposed to communicate. True False

1 answer:

xeze [42]3 years ago

7 0

Answer:

True

Explanation:

This are acts or actions that concur with situational expectations.

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A solution of pH 5 is diluted 100 times. Find the pH of the resulting solution.

liq [111]

First, we calculate of the concentration of the H+ ions in the solution from the pH given. Then, calculate the new concentration after dilution. Calculation are as follows:

pH = -log[H+]
5 = -log[H+]
[H+] = 1 x 10^-5 M

M1V1 = M2V2
<span>1 x 10^-5 M (V1) = M2(100V1)
</span>M2 = 1 x 10^-7

pH = -log[<span>1 x 10^-7</span>]
pH = 7

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3 years ago

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What statement would beast describe a digital signal

Assoli18 [71]

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3 years ago

What is the gram formula of na2co3

makkiz [27]

Answer:

106 gfm

Explanation:

element. number masses

Na. 2 23×2

C. 1. 12

O. 3. 16×3

then add

106 gfm

5 0

2 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

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The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

3 years ago