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ad-work [718]
4 years ago
8

Helppppp quick 25 points

Chemistry
2 answers:
KengaRu [80]4 years ago
8 0

Answer:

a

Explanation:

Vaselesa [24]4 years ago
5 0

Answer:

its A i just took the test

Explanation:

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When an irregularly shaped chunk of an unknown metal with a mass of 25.32 g was placed in a graduated cylinder containing 25.00
inn [45]

Answer: The density of the unknown metal is 7.86 g/ml.

Explanation:

Density is defined as the mass contained per unit volume.

Density=\frac{mass}{Volume}

Given : Mass of metal  = 25.32 g

Volume of metal = volume of water displaced = (28.22 - 25.00) ml = 3.22 ml

Putting in the values we get:

density=\frac{25.32g}{3.22ml}

Density=7.86g/ml

Thus the density of the unknown metal is 7.86 g/ml

7 0
4 years ago
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

brainly.com/question/20305871

#SPJ1

5 0
2 years ago
Si Units Worksheet. Use si units to complete
Keith_Richards [23]

\\ \bull\tt\dashrightarrow 2000m=2km

\\ \bull\tt\dashrightarrow 9000mL=9L

\\ \bull\tt\dashrightarrow 6L=6000mL

\\ \bull\tt\dashrightarrow 6cm=60mm

\\ \bull\tt\dashrightarrow 1000m=1km

\\ \bull\tt\dashrightarrow 50mm=5cm

\\ \bull\tt\dashrightarrow 4L=4000mL

\\ \bull\tt\dashrightarrow 10000g=10kg

\\ \bull\tt\dashrightarrow 10000m=10km

\\ \bull\tt\dashrightarrow 4000g=4kg

\\ \bull\tt\dashrightarrow 9km=9000m

\\ \bull\tt\dashrightarrow 3kg=3000g

\\ \bull\tt\dashrightarrow 90mm=9cm

\\ \bull\tt\dashrightarrow 4km=4000m

\\ \bull\tt\dashrightarrow 2000g=2kg

\\ \bull\tt\dashrightarrow 400cm=4m

\\ \bull\tt\dashrightarrow 3km=3000m

\\ \bull\tt\dashrightarrow 2cm=20mm

\\ \bull\tt\dashrightarrow 9000g=9kg

8 0
3 years ago
A ball is rolling across the floor. Every 3 seconds it travels 12 meters. Its speed, therefore, is 36 m/s. Is this problem solve
kirill [66]
No because you are supposed to do this

d = t)v
T for time and V for velocity or acceleration so what you do is take the two numbers

D= 3/12 it goes 12 meters every 3 seconds so you devide 12/3
12 \div 3 = 4
So for every second the ball roles 4 meters!
Hope this helps! :)
7 0
3 years ago
What is the difference between pure and applied chemistry?
bekas [8.4K]

Answer:

Explanation:

The major difference between pure and applied chemistry is the purpose and intent of the study.

Pure chemistry deals with the study of matter, matter transformations, and interactions between the different materials of the world, for only the sake of gaining empirical knowledge about the various substances that exist in the world. It does not really seek to apply this knowledge to do anything industrial.

Applied chemistry is the study of chemistry with the aim of utilizing this knowledge to solve the various problems that man faces. This approach of study is not for knowledge sake alone, rather it is for industrial application

3 0
3 years ago
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