What is radium used for

Need help plz asap .......

olganol [36]

Answer:

Purple flowers

Explanation:

Usually the dominant allele is a capital letter.

From the question, purple flowers are dominant to white flowers while white are recessive.

PP = Purple

pp = White

4 0

2 years ago

Read 2 more answers

HELP ME OUT PLS!!!!!!!

fgiga [73]

Answer:

the spiral type of galaxy

5 0

2 years ago

Read 2 more answers

Calculate the density of the four solutions. All of the solutions has the volume equal to 25 ml. Red solution has 25.0 g of mass

Scilla [17]

Answer:

B. Green solution density is 1.06 g/ml and blue solution density is 1.20 g/ml

Explanation:

Density is given as,

D = Mass / Volume

Red Solution,

D = 25 g / 25 mL

D = 1 g/mL

Green Solution,

D = 26.5 g / 25 mL

D = 1.06 g/mL

Yellow Solution,

D = 28.2 g / 25 mL

D = 1.128 g/mL

Blue Solution,

D = 30 g / 25 mL

D = 1.20 g/mL

3 0

2 years ago

An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr

Zina [86]

<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

$C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa$

Rate of flow of ideal gas , n = 4 kmol/hr = $\frac{4\times 1000mol}{3600s}=1.11mol/s$ (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)

We know that:

$\Delta U=0$ (For isothermal process)

So, by applying first law of thermodynamics:

$\Delta U=\Delta q-\Delta W$

$\Delta q=\Delta W$ .......(1)

Now, calculating the work done for isothermal process, we use the equation:

$\Delta W=nRT\ln (\frac{P_1}{P_2})$

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = final pressure = 50 kPa

Putting values in above equation, we get:

$\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW$

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

$\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW$

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

4 0

3 years ago

QUESTION 10

iris [78.8K]

Answer:

B

Explanation:

Phosphofructokinase-1

6 0

2 years ago