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2 years ago

What do punnet squares do

Chemistry

1 answer:

larisa86 [58]2 years ago

6 0

Answer:

The Punnett square is a square diagram that is used to predict the genotypes of a particular cross or breeding experiment. ... The diagram is used by biologists to determine the probability of an offspring having a particular genotype.

Explanation:

in biology it predicts the possible offspring provided the features of the mother or father

You might be interested in

Ch4+202==>c02+302 what type of chemical reaction

n200080 [17]

Answer:

First of all, the equation is typed wrong so it can easily be misinterpreted

Ethane (CH4) + Oxygen gas (O2) would give us Carbon Dioxide (CO2) and WATER (H2O)

CH4 + 2O2 -----> CO2 + 2H2O

And this is a combustion reaction since we have oxygen as a reactant and carbon dioxide and water as products.

4 0

2 years ago

Students asked to measure the temperature of a reaction beaker recorded the following temperatures. 103.7°C, 108.4°C, 105.8°C, 1

Nata [24]

There are two terms

a) accuracy : it relates to the exactness of an answer that how an answer is close to the actual answer or actual reading

So 104.6 is accurate

b) Precision : This is related to the closeness of different readings with each other

The first reading is 103.7 and the second one is 108.4 so the second reading is quite different from the first reading so it cannot be called as precised

Again 105.8 has good difference from the second reading hence again this is not precised

However the last reading 104.6 is quite near to 105.8 so 104.6 can be called as precise

6 0

3 years ago

Read 2 more answers

An equilibrium mixture of CO, O2 and CO2 at a certain temperature contains 0.0010 M CO2 and 0.0025 M O2. At this temperature, Kc

pantera1 [17]

Answer:

$5.35 *10^{-4}$ M

Explanation:

Equation for the reaction is as follows:

$2CO_{(g)}$ + $O_{2(g)}$ ⇄ $2CO_{2(g)}$

By Applying the ICE Table; we have

$2CO_{(g)}$ + $O_{2(g)}$ ⇄ $2CO_{2(g)}$

Initial x 0.0025 M 0.0010 M

Change 0 0 0

Equilibrium x 0.0025 M 0.0010 M

$K_c =\frac{[CO_2]^2}{[CO]^2[O_2]}$

Given that $K_c = 1.4*10^2$ ; Then:

$1.4 *10^2 = \frac{(0.001)^2}{(x)^2(0.025)}$

$1.4 *10^2*0.025 = \frac{(0.001)^2}{(x)^2}$

$3.5 =( \frac{(0.001)}{(x)})^2$

$\sqrt {3.5} = \sqrt {( \frac{(0.001)}{(x)} )^2}$

$1.87=\frac{(0.001)}{(x)}$

$(x)= \frac{(0.001)}{1.87 }$

$x = 5.35 *10^{-4}$ M

∴ The equilibrium concentration of CO = $x = 5.35 *10^{-4}$ M

7 0

3 years ago

Which of the following represents the least number of molecules?

lozanna [386]

Answer:

A ; 20g of water has the least number of molecules

Explanation:

Here, we want to know which of the options represent the least number of molecules;

To calculate this, we are going to first calculate the number of moles using the formula below;

Number of moles = mass/molar mass

Then from the number of moles, we can calculate the number of molecules by multiplying the number of moles by 6.02 * 10^23 molecules

Let’s proceed;

a. That will be 20/18.02 = 1.1099 moles

1.1099 * 6.02 * 10^23 = 6.68 * 10^23 molecules

b. That will be 77/16.06 = 4.8 moles

4.8 * 6.02 * 10^23 = 2.89 * 10*24 molecules

c. That will be 68/42.09 = 1.62 moles

1.62 * 6.02 * 10^23 = 9.73 * 10^23 molecules

d. That will be 100/44.02 = 2.27 moles

2.27 * 6.02 * 10^23 = 1.37 * 10^24 molecules

e. That will be = 84/20.01 = 4.2 moles

4.2 * 6.02 * 10^23 = 2.53 * 10^24 molecules

From all the values obtained, the lowest is 20g of water

6 0

3 years ago

g The combustion of 1.877 1.877 g of glucose, C 6 H 12 O 6 ( s ) C6H12O6(s), in a bomb calorimeter with a heat capacity of 4.30

Mekhanik [1.2K]

Answer:

-2.80 × 10³ kJ/mol

Explanation:

According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qcal) and the heat released by the combustion of the glucose (Qcomb) is zero.

Qcal + Qcomb = 0

Qcomb = - Qcal [1]

We can calculate the heat absorbed by the bomb calorimeter using the following expression.

Qcal = C × ΔT = 4.30 kJ/°C × (29.51°C - 22.71°C) = 29.2 kJ

where,

C: heat capacity of the calorimeter

ΔT: change in the temperature

From [1],

Qcomb = - Qcal = -29.2 kJ

The internal energy change (ΔU), for the combustion of 1.877 g of glucose (MW 180.16 g/mol) is:

ΔU = -29.2 kJ/1.877 g × 180.16 g/mol = -2.80 × 10³ kJ/mol

3 0

3 years ago