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3 years ago

Convert 55mph to m/s..

Chemistry

1 answer:

andrezito [222]3 years ago

5 0

24.5872 this iwhat i got

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Balance the following equation. Choose "blank" if no coefficient other than 1 is needed.

Ann [662]

Explanation:

Pb+2 AgNO3=Pb(NO3)2+2Ag

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3 years ago

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Calculate the amount of heat needed to boil 64.7 g of benzene (C6H6), beginning from a temperature of 41.9 C . Round your answer

JulsSmile [24]

Answer: The amount of heat needed is = 4.3kJ

Explanation:

Amount of heat H = M × C × ΔT

M= mass of benzene = 64.7g

C= specific heat capacity = 1.74J/gK

ΔT = T2-T1

Where T1 is initai temperature = 41.9C

T2 is the final temperature( boiling point of benzene) = 80.1C

H= 64.7×1.74×80.7

H= 4300J

H=4.3kJ

Therefore, the amount of heat needed is 4.3kJ

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3 years ago

Marina CMI [18]

Answer:

Molar mass = 30.026 g/mol

Explanation:

Let Molar mass be mm

mm of CH2O = 12.01 + 2(1.008) + 16.00

= 12.01 + 2.016 + 16.00 = 30.026 g/mol

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2 years ago

PLEASE HELP

Sladkaya [172]

Answer:

C) rusting ability

Explanation:

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3 years ago

what is the pH of a solution that results when 0.010mol HNO3 is added to 500.ml of a solution that is 0.10M in aqueous ammonia a

WARRIOR [948]

Answer : The

pH of a solution is, 8.56

Explanation : Given,

$K_b=1.8\times 10^{-5}$

Concentration of ammonia (base) = 0.10 M

Concentration of ammonium nitrate (salt) = 0.55 M

First we have to calculate the value of $pK_b$ .

The expression used for the calculation of $pK_b$ is,

$pK_b=-\log (K_b)$

Now put the value of $K_b$ in this expression, we get:

$pK_b=-\log (1.8\times 10^{-5})$

$pK_b=5-\log (1.8)$

$pK_b=4.7$

Now we have to calculate the pOH of buffer.

Using Henderson Hesselbach equation :

$pOH=pK_b+\log \frac{[Salt]}{[Base]}$

Now put all the given values in this expression, we get:

$pOH=4.7+\log (\frac{0.55}{0.10})$

$pOH=5.44$

The pOH of buffer is 5.44

Now we have to calculate the pH of a solution.

$pH+pOH=14\\\\pH+5.44=14\\\\pH=14-5.44\\\\pH=8.56$

Thus, the pH of a solution is, 8.56

8 0

3 years ago