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fomenos
3 years ago
7

A ball is kicked straight up. The ball's speed decreases as it moves upward. Why does the ball's

Chemistry
1 answer:
kumpel [21]3 years ago
8 0

Answer:

I think the third one

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The rate constant for the first-order decomposition of n2o is 3.40 s-1. what is the half-life of the decomposition?
stiks02 [169]

Hey There:

First order  half life equation :

T 1/2  = ln ( 2 ) /K

T 1/2 = 0.693 / 3.40

T 1/2 = 0.204 s

Answer B

3 0
3 years ago
Read 2 more answers
BRAINLIESTTT ASAP!! PLEASE HELP ME :)
a_sh-v [17]

Answer: true

Explanation:

3 0
3 years ago
When 47.1 J of heat is added to 14.0 g of a liquid, its temperature rises by 1.80 ∘C. What is the heat capacity of the liquid?
Alja [10]

Answer:

\boxed {\boxed {\sf 1.87 \J/g \textdegree C}}

Explanation:

We are asked to find the specific heat capacity of a liquid. We are given the heat added, the mass, and the change in temperature, so we will use the following formula.

q= mc\Delta T

The heat added (q) is 47.1 Joules. The mass (m) of the liquid is 14.0 grams. The specific heat (c) is unknown. The change in temperature (ΔT) is 1.80 °C.

  • q= 47.1 J
  • m= 14.0 g
  • ΔT= 1.80 °C

Substitute these values into the formula.

47.1 \ J = (14.0 \ g) * c * (1.80 \textdegree C)

Multiply the 2 numbers in parentheses on the right side of the equation.

47.1 \ J = (14.0 \ g * 1.80 \textdegree C)*c

47.1 \ J = (25.2 \ g*\textdegree C) *c

We are solving for the heat capacity of the liquid, so we must isolate the variable c. It is being multiplied by 25.2 grams * degrees Celsius. The inverse operation of multiplication is division, so we divide both sides of the equation by (25.2 g * °C).

\frac {47.1 \ J}{(25.2 g *\textdegree C)} = \frac {(25.2 g *\textdegree C)*c}{{(25.2 g *\textdegree C)}}

\frac {47.1 \ J}{(25.2 g *\textdegree C)} =c

1.869047619 \ J/g *\textdegree C = c

The original measurements of heat, mass, and temperature all have 3 significant figures, so our answer must have the same. For the number we found that is the hundredth place. The 9 in the thousandth place to the right tells us to round the 6 up to a 7.

1.87 \ J/ g * \textdegree C =c

The heat capacity of the liquid is approximately 1.87 J/g°C.

3 0
3 years ago
a sample of gold required 2.1200j of heat to melt it from room temperature, 22.0 degrees celsius to its melting point, 1064.4 de
ElenaW [278]

Mass of Gold = 267.165 ×  0.01552494829

⇒ 4.1477228099

The amount of heat(q) required to raise m grams of a substance-specific C from T1 to T2 is given by

q=m C (T2-T1) ........1

Given : q= 2.1200 J

the initial temperature of gold, T1 = 22.0Celcius

the final temperature of gold, T2 = 1064.4Celcius

specific heat of gold = 0.131

putting values in eq 1:

⇒ 2.1200 = m × 0.131 × (1064.4-22)

⇒ 2.1200 = m × 0.131 × 1042.4

⇒ 2.1200 / 136.5544

⇒ 0.01552494829

Since 1g= 0.01552494829 Pounds

Mass of Gold = 267.165 ×  0.01552494829

⇒ 4.1477228099

Learn more about temperature here: brainly.com/question/11464844

#SPJ9

7 0
1 year ago
Three isotopes of argon occur in nature – 36 18Ar, 38 18Ar, 40 18Ar. Calculate the average atomic mass of argon to two decimal p
likoan [24]

Answer: 3) 39.96 amu

Explanation:

Mass of isotope Ar- 36 = 35.97 amu

% abundance of isotope Ar- 36= 0.337% = \frac{0.337}{100}=3.37\times 10^{-3}

Mass of isotope Ar- 38 = 37.96 amu

% abundance of isotope 2 = 0.063 % = \frac{0.063}{100}=6.3\times 10^{-4}

Mass of isotope Ar- 40 = 39.96 amu

% abundance of isotope 2 = 99.600 % = \frac{99.600}{100}=0.996

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(35.97\times 3.37\times 10^{-3})+(37.96\times 6.3\times 10^{-4})+(39.96\times 0.996)]

A=39.96amu

Therefore, the average atomic mass of argon is 39.96 amu

4 0
4 years ago
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