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3 years ago

A large gently sloping volcano with a fluid eruption

1 answer:

6 0

Salutations!

A large gently sloping volcano with a fluid eruption is called shield volcano. The fluid is basaltic lava and the slops are gently sliding.

Hope I helped (:

Have a great day!

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The only pair that would result in a recessive phenotype is:

Inessa05 [86]

I believe cc would result in a recessive pair.

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2 years ago

Read 2 more answers

Rasheed calculates that his chemical reaction should produce 4 moles of product, but when he does the experiment, he gets only 3

Brut [27]

Answer:

The answer to your question is 75%

Explanation:

Data

Theoretical production = 4 moles

Experimental production = 3 moles

Percent yield = ?

Formula

$Percent yield = \frac{Experimental production}{Theoretical production} x 100$

Substitution

$Percent yield = \frac{3}{4} x 100$

Result

Percent yield = 75 %

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3 years ago

Arthur is performing a lab which requires a 0.2 M solution of calcium hydroxide. However, the only calcium hydroxide solution av

aev [14]

Answer:

He needs to add 4 mL of the 0.5 M solution to 6 mL of water.

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3 years ago

what kind of generalization can you make about how the number of protons and neutrons are related to each other in the elements

AURORKA [14]

... to be called elements. This lesson shows you how to predict the numbers of neutrons, electrons, andprotons of the isotopes they are likely to find in nature. (cont.) ... What kind of generalization can you make about how the number of protons and neutrons are related to each other in the elements? Unit 1 • Investigation IV

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3 years ago

15.5 g of an unknown metal at 165.0°C is dropped into 150.0mL of H2O at 23.0°C in a coffee cup calorimeter. The metal and H2O re

tino4ka555 [31]

Answer:

Specific heat capacity of metal is 2.09 j/g.°C.

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Given data:

Mass of metal = 15.5 g

Initial temperature = 165.0°C

Initial temperature of water = 23.0°C

Final temperature = 30.0°C

Specific heat capacity of metal = ?

Specific heat capacity of water = 4.184 J/g°C

Volume of water = 150.0 mL or 150.0 g

Solution:

Formula:

- Qm = +Qw

Now we will put the values in formula.

-15.5 g × c × [ 30.0°C - 165.0°C] = 150 g × 4.184 J/g°C × [ 30.0°C - 23.0°C]

15.5 g × c × 135°C = 4393.2 j

2092.5 g.°C × c = 4393.2 j

c = 4393.2 j/2092.5 g.°C

c = 2.09 j/g.°C

4 0

3 years ago