Ask question

Ask question

All categories

4 years ago

12

Which type of allergy test, also called a subcutaneous test, introduces dilute solutions of allergens into the skin of the inner

forearm or upper back with a fine-gauge needle?

1 answer:

alexgriva [62]4 years ago

6 0

Answer:

The type of allergy test is scratch test.

Explanation:

In a scratch test also called a skin pri c k test or puncture test, an allergy provoking substance placed on the skin is gradually scratched into the skin surface with the aid of a needle or lancet so as to detect bodily reactions to allergens. Some of the types allergens whose reactions are generally tested for include pollen, dust mites, molds and foods.

The scratch test takes about 5 to 10 minutes and it could cause slight pain and the may result in itching feeling at the location where the allergen was introduced.

You might be interested in

Consider the reaction: A (aq) ⇌ B (aq) at 253 K where the initial concentration of A = 1.00 M and the initial concentration of B

Ugo [173]

Answer:

The maximum amount of work that can be done by this system is -2.71 kJ/mol

Explanation:

Maximum amount of work denoted change in gibbs free energy $(\Delta G)$ during the reaction.

Equilibrium concentration of B = 0.357 M

So equilibrium concentration of A = (1-0.357) M = 0.643 M

So equilibrium constant at 253 K, $K_{eq}= \frac{[B]}{[A]}$

[A] and [B] represent equilibrium concentrations

$K_{eq}=\frac{0.357}{0.643}=0.555$

When concentration of A = 0.867 M then B = (1-0.867) M = 0.133 M

So reaction quotient at this situation, $Q=\frac{0.133}{0.867}=0.153$

We know, $\Delta G=RTln(\frac{Q}{K_{eq}})$

where R is gas constant and T is temperature in kelvin

Here R is 8.314 J/(mol.K), T is 253 K, Q is 0.153 and $K_{eq}$ is 0.555

So, $\Delta G=8.314\times 253\times ln(\frac{0.153}{0.555})mol/K$

= -2710 J/mol

= -2.71 kJ/mol

4 0

3 years ago

Which alkane would have a higher boiling point?a)ethaneb)pentanec)heptane

Whitepunk [10]

Answer:

The correct answer is option c.

Explanation:

Alkanes with higher molecular mass has higher boiling point.

Thisis because when the molar mass of the alkanes increases the the surface area increases with which van der Waals forces between the molecules of alkane also increase which increases the association of the molecules of with each other which results in increase in boiling point is observed.

The increasing order of the molar mass of the given alkanes;

$C_2H_6$

So out of ethane, pentane and heptane . Heptane has highest molecular mass with higher boiling point value. Where as ethane have the lowest value of boiling point.

6 0

4 years ago

Four atoms are arbitrarily labeled D, E, F, and G. Their electronegativities are as follows: D = 3.8, E = 3.3, F = 2.8, and G =

PSYCHO15rus [73]

Answer:

Thus the order of covalent character will be

DG < EG < DF < DE

Explanation:

The electronegativity decides the nature of bond formed between two atoms.

More the difference in electronegativity more the ionic character in the bond formed.

Let us check the electronegativity difference between the elements forming molecules

a) DE :

3.8-3.3 = 0.5

b) DG :

3.8-1.3 = 2.5

c) EG

3.3-1.3 = 2

d) DF

3.8-2.8 = 1.3

So the maximum ionic character will be in DG and minimum in DE

Thus the order of covalent character will be

DG < EG < DF < DE

3 0

4 years ago

The sulfur dioxide (so2) stack-gas concentration from fossil-fuel combustion is 12 ppmv. determine the stack-gas so2 concentrati

sergejj [24]

The concentration of $SO_{2}$ in the stack gas = 12 ppmv

That means 12 L of $SO_{2}$ is present per $10^{6} L gas$

The given temperature is 273 K (0 C) and pressure is 1 atm. At these conditions, 1 mol of gas would occupy,

$PV = nRT$

$(1 atm) (V) = (1 mol)(0.08206\frac{L.atm}{mol.K}) (273 K)$

V = 22.4 L

1 mol $SO_{2}$ occupies 22.4 L

Moles of $SO_{2}$ = $12 L *\frac{1 mol}{22.4 L} = 0.5357 mol SO_{2}$

Mass of $SO_{2}$ = $0.5357 mol *\frac{64.06 g}{1 mol} = 34.32 g SO_{2} *\frac{10^{6} microgram}{1 g}$ = $3.432 *10^{7}$ μg

Converting $10^{6} L to m^{3}$ :

$10^{6} L *\frac{1 m^{3}}{1000 L}$ = $10^{3} m^{3}$

Calculating the concentration in μg/ $m^{3}$ :

$\frac{3.432 * 10^{7} microgram}{10^{3} L} = 3.432 * 10^{4} microgram/m^{3}$

3 0

3 years ago

If 12% of x is equal to 6% of y, then 18% of x will be equal to how much percent of y?

ipn [44]

18 percent of x will be equal to 12 percent of y

7 0

3 years ago