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3 years ago

During a medical screening, lung capacity testing is a standard procedure. true or false

Chemistry

2 answers:

grin007 [14]3 years ago

8 1

I believe it is false

best of luck hope this helps
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Naya [18.7K]3 years ago

6 0

Answer;

-The above statement is true.

During a medical screening, lung capacity testing is a standard procedure

Explanation;

-Lung capacity testing also the pulmonary function test is a part of medical screening that is done to diagnose certain types of lung disease, such as asthma, bronchitis, and emphysema. Find the cause of shortness of breath and also measure whether exposure to chemicals at work affects lung function.

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vazorg [7]

Answer:

characteristics i believe

8 0

2 years ago

Read 2 more answers

Sierra did 500 J of work to move her couch. If she exerts 250 N of force on the couch, How fat did she move it? (Work: W = Fd

Elodia [21]

W=F*d

W= 500 J
F = 250 N

500 J = 250 N * d
d= 500J/250 N = 2 J/N = 2(N*m)/N = 2 m
Answer is 2 m.

7 0

3 years ago

An object has a mass of 3.50 grams and a density of 5.61 g/mL. What is the volume of this object? *

Elina [12.6K]

Answer:

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density} \\$

From the question we have

$volume = \frac{3.50}{5.61} \\ = 0.6238859...$

We have the final answer as

Hope this helps you

5 0

2 years ago

The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t

slava [35]

<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol$

The given chemical equation follows:

$2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4$

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = $\frac{1}{2}\times 0.133=0.0665moles$ of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

$0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g$

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

8 0

3 years ago

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

2 years ago