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BlackZzzverrR [31]
3 years ago
8

What type of chart is used to help organize study and predict genetic inheritance? ​

Chemistry
2 answers:
Naily [24]3 years ago
8 0

Answer:

Its a punnet square

Explanation:

Oliga [24]3 years ago
4 0
The type of chart is a graph
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What information is needed to determine the amount of moles in 1.6g of HCL?
alukav5142 [94]

Answer:

Molar mass and Mass

Explanation:

The relationship between mass and number of moles is given as;

Number of moles = Mass / Molar mass

Mass = 1.6 g

Molar mass of HCl = ( 1 + 35.5 ) = 36.5 g/mol

Number of moles = 1.6 g / 36.5 g/mol

Number of moles = 0.0438 mol

8 0
3 years ago
How many grams S03<br> are needed to make 400g<br> H₂So4 in 49%
NemiM [27]

160 g of SO3 are needed to make 400 g of 49% H2SO4.

<h3>How many grams of SO3 are required to prepare 400 g of 49% H2SO4?</h3>

The equation of the reaction for the formation of H2SO4 from SO3 is given below as follows:

SO_{3} + H_{2}O \rightarrow H_{2}SO_{4}

1 mole of SO3 produces 1 mole of H2SO4

Molar mass of SO3 = 80 g/mol

Molar mass of H2SO4 = 98 g/mol

80 g of SO3 are required to produce 98 og 100%H2SO4

mass of SO3 required to produce 400 g of 100 %H2SO4 = 80/98 × 400 = 326.5 g of SO3

Mass of SO3 required to produce 49% of 400 g H2SO4 = 326.5 × 49% = 160 g

Therefore, 160 g of SO3 are needed to make 400 g of 49% H2SO4.

Learn more about mass and moles at: brainly.com/question/15374113

#SPJ1

7 0
2 years ago
What is the smallest part of an element?
Ainat [17]
An atom is the smallest part of an element
4 0
3 years ago
Read 2 more answers
Please help me!!!!!!!!!!
ladessa [460]

Answer:

i can't see the image so sorry

Explanation:

i couldnt see the imagee

5 0
3 years ago
F
Airida [17]

Answer:

Final pressure = 362.7 Pa

Explanation:

Given that,

Initial volume, V₁ = 930 ml

Initial pressure P₁ = 156 Pa

Final volume, V₂ = 400 mL

We need to find the final pressure. We know that the relation between volume and pressure is inverse i.e.

V\propto \dfrac{1}{P}\\\\\dfrac{V_1}{V_2}=\dfrac{P_2}{P_1}\\\\P_2=\dfrac{V_1P_1}{V_2}\\\\P_2=\dfrac{930\times 156}{400}\\\\P_2=362.7\ Pa

So, the final pressure is equal to 362.7 Pa.

3 0
3 years ago
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