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maksim [4K]
4 years ago
14

How many moles is 2.80 x 10^24 atoms of silicon

Chemistry
1 answer:
neonofarm [45]4 years ago
3 0
The answer that i found is 4.650
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Lipid-soluble hormones can't be stored within vesicles.<br> Select one:<br> O True<br> O False
Darya [45]

Answer:

true

Explanation:

they can not be stored within the vesicles

7 0
2 years ago
A 100. g iron rod is heated to an unknown initial temperature. While cooling down to a temperature of 25.0 degree C, the rod rel
AfilCa [17]

Answer: a)  38.3^0C

Explanation:

As we know that,  

q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

where,

q = heat released  = -600.2 J

m = mass of iron rod =100 g  

T_{final} = final temperature =25.^0C

T_{initial} = initial temperature = ?

c = specific heat of iron = 0.452J/g^0C

Now put all the given values in equation, we get

-600.2J=100\times 0.452J/g^0C\times (25.0-T_{initial})^0C]

T_{initial}=38.3^0C

Therefore, the initial temperature of iron rod was 38.3^0C

6 0
3 years ago
Easy chemistry!! I'm really on a time crunch please please please help me!!!! I'm terrible at chemistry x﹏x
Serjik [45]

Answer:

ADD MORE POINTS THEN WE CAN TALK HUN

Explanation:

YW!!!!! ;D

8 0
3 years ago
The specific heats and densities of several materials are given below:
krek1111 [17]

<u>Answer:</u> The change in temperature is 84.7°C

<u>Explanation:</u>

To calculate the change in temperature, we use the equation:

q=mc\Delta T

where,

q = heat absorbed = 1 kCal = 1000 Cal    (Conversion factor: 1 kCal = 1000 Cal)

m = mass of steel = 100 g

c = specific heat capacity of steel = 0.118 Cal/g.°C

\Delta T = change in temperature = ?

Putting values in above equation, we get:

1000Cal=100g\times 0.118Cal/g.^oC\times \Delta T\\\\\Delta T=\frac{1000}{100\times 0.118}=84.7^oC

Hence, the change in temperature is 84.7°C

6 0
3 years ago
The sum of the first 10 terms of an arithmetic progression is 120 and the sum of first twenty is 840. find sum of first 30 terms
STatiana [176]

Answer:

The sum of first 30 terms of the arithmetic progression is <u>2160.</u>

Explanation:

For an arithmetic progression, the sum of first n terms with first term as a and common difference d is given as:

S_n=\frac{n}{2}(2a+(n-1)d)

Now, it is given that:

For\ n=10,S_n=120\\For\ n=20,S_n=840

Now, plug in these values and frame two equations in a\ and\ d

S_{10}=\frac{10}{2}(2a+(10-1)d)\\120=5(2a+9d)\\2a+9d=\frac{120}{5}\\2a+9d=24------------1

S_{20}=\frac{20}{2}(2a+(20-1)d)\\840=10(2a+19d)\\2a+19d=\frac{840}{10}\\2a+19d=84-----------2

Now, we solve equations (1) and (2) for a\ and\ d. Subtract equation (1) from equation (2). This gives,

2a+19d-2a-9d=84-24\\19d-9d=60\\10d=60\\d=\frac{60}{10}=6

Now, plug in the value of d=6 in equation (1) and solve for a.

2a+9(6)=24\\2a+54=24\\2a=24-54\\2a=-30\\a=\frac{-30}{2}=-15

Plug in the values of a=-15,\ n=30\ and\ d=6 in the sum formula to find the sum of first 30 terms.

Now, the sum of first 30 terms is given as:

S_{30}=\frac{30}{2}(2(-15)+(30-1)(6))\\S_{30}=15(-30+29(6))\\S_{30}=15(-30+174)\\S_{30}=15(144)=2160

Therefore, the sum of first 30 terms of the arithmetic progression is 2160.

4 0
3 years ago
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