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Ksju [112]
3 years ago
13

For years, surgeons have had great success using _____.

Chemistry
2 answers:
sweet-ann [11.9K]3 years ago
7 0
C artificial heart valves
FrozenT [24]3 years ago
6 0
C the artifical heart
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Convert 3.8 Km/sec to miles/year
weqwewe [10]
We are asked to convert from units of kilometer per second to units of miles per year. To do this, we need a conversion factor which would relate the different units involved. We either multiply or divide this certain value to the original measurement depending on what is asked. From literature, we will find that 1 mile is equal to 1609 meters and 1000 m is equal to 1 kilometer. Also, we will find that 3600 s is equal to 1 hr, 24 hr is equal to 1 day and 365 days is equal to 1 year. We do the conversion as follows:

3.8 km / s ( 1000 m / 1 km ) ( 1 mile / 1609 meters ) ( 3600 s / 1 hr ) ( 24 hr / 1 day ) ( 365 days / 1 year ) = 74479055.3 miles per year
5 0
3 years ago
Which three elements affect air pressure?
Sloan [31]
A sorry if it’s wrong good luck
8 0
3 years ago
A compound contains 34.5% calcium, 24.1% silicon and 41.4% oxygen by mass. What is its empirical formula?
Amiraneli [1.4K]

<em>empirical \: formula \\  = CaSiO3 \\ please \: see \: the \: attached \: picture \\ hope \: it \: helps</em>

4 0
3 years ago
Which of the following has the largest % of sodium, 20g of sodium fluoride, 20g of sodium chloride, or 20g of sodium bromide?
Ksenya-84 [330]
The one whose molar mass is smaller than sodium. 

Fluorine is 19 g
Chlorine= 35.5 g
Bromine= 80 g

so in sodium fluoride, the percent of sodium is the largest. 
4 0
4 years ago
How many kilocalories are required to increase the temperature of 15.6 g of iron from 122 °c to 355 °c. the specific heat of iro
Dmitriy789 [7]

Heat require to boil 15.6 g iron from 122 C0to 355 C0 whereas,

Q = m s dT

Where, m is mass of iron

s is specific heat of iron

d T is change in temperature in celcius

= 15.6 g * 0.45 J /g /C * (355 - 122)  = 1.63 * 10^3 J

If  

1 cal = 4.2 J

Then,  

Q = (1.63 * 10^3) /4.2 = 0.389 K cal

Thus 0.389 k cal of enrgy  is required by a 15.6 g Fe to reach to 355 C^0

4 0
4 years ago
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