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Nataliya [291]
3 years ago
11

Convert 0.00590 lb/qt to g/gal Please help me i will mark the brainliest!!

Chemistry
1 answer:
Dima020 [189]3 years ago
8 0
According to google, one gallon equals 8.36 pounds. I took 8.36and divided by 0.00590 and got 1416.94915254 So if rounded to the nearest hundreth, 1416.95. I have no idea if I'm right. Sorry I couldn't do more!
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Cho 4g CuO vào dung dịch axit clohidric 10% thì phản ứng vừa đủ.
Sholpan [36]

Answer:

Explanation:

a. CuO+ 2HCl⇒CuCl2+ H2O

b. n_{CuO}= \frac{4}{80}= 0,05 (mol)

⇒n_{CuCl2}= n_{CuO}=0,05 mol

⇒m_{CuCl2}= 0,05×135=6,75 (g)

c. n_{HCl}=2× n_{CuO}=0,1 (mol)

⇒m_{HCl}= 0,1×36,5= 3,65 (g)

⇒m_{dd HCl}= \frac{m_{HCl}}{10}×100=36,5 (g)

⇒ Nồng độ phần trăm dd sau phản ứng= Nồng độ % dd CuCl2=\frac{m_{CuCl2} }{m_{dd HCl+ m_{CuO} } }×100=\frac{6,75}{36,5+4} ×100≈ 16,67%

8 0
3 years ago
Which choice below would match homozygous recessive?
LUCKY_DIMON [66]
The answer is B

Homozygous mean the same and you know that the two alleles will be the same (either BB or bb) and receive is usually the lower case set of alleles
8 0
2 years ago
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What is the mass of KNO3 should be used to prepare 2.00 L of a 0.700 M solution?
Dmitry [639]

Answer:

141.4g

Explanation:

Moles= conc. ×vol

m/M = 0.7×2

m/101 = 0.7×2

m= 141.4g

3 0
3 years ago
A sample that weighs 107.75 g is a mixture of 30% helium atoms and 70% krypton atoms. How many particles are present in the samp
r-ruslan [8.4K]
<span>A sample that weighs 107.75 g is a mixture of 30% helium atoms and 70% krypton atoms. How many particles are present in the sample?    </span>3.59
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3 years ago
Nitrogen dioxide (NO2) cannot be obtained in a pure form in the gas phase because it exists as a mixture of NO2 and N2O4. At 16°
Pavel [41]

Answer:

PNO₂ = 0.49 atm

PN₂O₄ = 0.45 atm

Explanation:

Let's begin with the equation of ideal gas, and derivate from it an equation that  involves the density (ρ = m/V).

PV = nRT

n = m/M (m is the mass, and M the molar mass)

PV = \frac{m}{M}RT

PxM = \frac{m}{V}RT

PxM = ρRT

ρ = PxM/RT

With the density of the gas mixture, we can calculate the average of molar mass (Mavg), with the constant of the gases R = 0.082 atm.L/mol.K, and T = 16 + 273 = 289 K

2.7 = \frac{0.94xMavg}{0.082x289}

0.94Mavg = 63.9846

Mavg = 68.0687 g/mol

The molar mass of N is 14 g/mol and of O is 16 g/mol, than M_{NO2} = 46 g/mol and M_{N2O4} = 96 g/mol. Calling y the molar fraction:

Mavg = M_{NO2}y_{NO2} + M_{N2O4}y_{N2O4}

And,

y_{NO2} + y_{N2O4} = 1

y_{N2O4} = 1 - y_{NO2}

So,

68.0687 = 46y_{NO2} + 92x(1 - y_{NO2})

68.0687 - 92 = 46y_{NO2} - 92y_{NO2}

46y_{NO2} = 23.9313

y_{NO2} = 0.52

y_{N2O4} = 0.48

The partial pressure is the molar fraction multiplied by the total pressure so:

PNO₂ = 0.52x0.94 = 0.49 atm

PN₂O₄ = 0.48x0.94 = 0.45 atm

8 0
3 years ago
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