A landslide A. weathering B. erosion C. deposition

What are three possible blood type alleles?

Keith_Richards [23]

Answer:

Three possible blood type alleles are Iᴬ, Iᴮ and i

Explanation:

Iᴬ, Iᴮ and i are three possible blood type alleles.

Iᴬ and Iᴮ are known as co-dominant, and The i allele is recessive.

Thus, Three possible blood type alleles are Iᴬ, Iᴮ and i

-TheUnknownScientist

8 0

2 years ago

Read 2 more answers

Write the balanced chemical equation for the reaction of glucose (C6,H12,O6) with oxygen gas to produce carbon dioxide gas

Fittoniya [83]

Answer:

C6H12O6+6O2=6Co2+6H2o

4 0

2 years ago

Determine all values of hh and k for which the system has no solution

solniwko [45]

Explain more so i could answer it!!

6 0

2 years ago

When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 20.4 g g of carbon were burned in the presence

Aleonysh [2.5K]

Answer:

C + O2 -> CO2

C 20.4/12 = 1.7

O 70.4-16.0/16=3.4

CO2 1.7 mol * 44 = 74.8

3 0

2 years ago

Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)

Brums [2.3K]

Answer:1) Volume of $AgNO_3$ required is 55.98 mL.

2) 0.62577 grams of $Ag_3PO_4$ is produced.

Explanation:

$3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)$

1) Molarity of $AgNO_3,M_1=0.225 M$

Volume of $AgNO_3.V_1=?$

Molarity of $Na_3PO_4,M_2=0.135 M$

Volume of $Na_3PO_4,V_2=31.1 mL=0.0311 L$

$Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

$\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles$

According to reaction, 1 mole of $Na_3PO_4$ reacts with 3 mole of $AgNO_3$ , then, 0.0041985 moles of $Na_3PO_4$ will react with:

$\frac{3}{1}\times 0.0041985$ moles of $AgNO_3$ that is 0.0125955 moles.

$M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}$

$V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL$

Volume of $AgNO_3$ required is 55.98 mL.

2)

$Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

Number of moles of $AgNO_3=0.195\times 0.023 L=0.004485 moles$

According to reaction, 3 moles of $AgNO_3$ gives 1 mole of $Ag_3PO_4$ , then 0.004485 moles of $AgNO_3$ will give: $\frac{1}{3}\times 0.004485$ moles of $Ag_3PO_4$ that is 0.001495 moles.

Mass of $Ag_3PO_4$ =

Moles of $Ag_3PO_4$ × Molar Mass of $Ag_3PO_4$

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of $Ag_3PO_4$ is produced.

7 0

3 years ago

A landslideA. weathering B. erosion C. deposition​

A landslideA. weathering B. erosion C. deposition