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3 years ago

Why are allergic reactions often

1 answer:

7 0

Allergic reactions (hypersensitivity reactions) are inappropriate responses of the immune system to a normally harmless substance

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A metallurgist is making an alloy that consists of 325 grams of chromium (Cr) and 2.5 kg of iron (Fe). Find the total mass of th

Darya [45]

Answer:- Mass of the alloy is 2.8 kg.

Solution:- Mass of Cr in the alloy is 325 g and mass of Fe in the alloy is 2.5 kg. Mass of alloy would be the sum of masses of constituent metals.

Masses of the metals are not in the same units. So, we need to make the units equal. The want answer in kg so let's convert mass of Cr from g to kg.

Since, 1000 g = 1 kg

So, $325g\frac{1kg}{1000g}$

= 0.325 kg

Mass of alloy = mass of Cr + mass of Fe

mass of alloy = 0.325 kg + 2.5 kg = 2.825 kg

If we consider significant figures then as per the rules, the answer should not have more than one decimal place.

So, 2.825 kg is round off to 2.8 kg and hence the mass of the alloy is 2.8 kg.

5 0

3 years ago

Are the atoms really "sharing" electrons? Explain.

Iteru [2.4K]

Answer:

yes, in certain cases

there are different types of bondings between atoms

and in some they lend electrons to make their atom stable this type of bonding is called ionic bonding

and in covalent bond the atoms share their electrons

7 0

3 years ago

Which of these is a characteristic of science? (5 points) Question 1 options: 1) It cannot be reproduced by any scientist. 2) It

Alex17521 [72]

Answer:

Explanation:

It is based on empirical evidence

4 0

3 years ago

Use this equation for the following problems: 2NaN3 --> 2Na+3N2

olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

2NaN₃ → 2Na + 3N₂,

It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

using cross multiplication:

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

Finally, we can get the grams of NaN₃ needed:

mass = no. of moles x molar mass = (0.517 mol)(65.0 g/mol) = 33.6 g.

Q2: How many mL of N₂ result if 8.3 g Na are also produced?

We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

We can get the no. of moles of N₂ produced with 0.36 mol of Na:

using cross multiplication:

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

We can get the mass of 0.54 mol of N₂:

mass = no. of moles x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

Now, we can get the mL of 15.12 g of N₂:

using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

∴ The volume of N₂ result = (1.0 L)(15.12 g)/(0.92 g) = 16.434 L = 16434 mL.

4 0

2 years ago

Una disolución contiene 40% de ácido acético en masa. La densidad de la disolución es de 1.049 g/mL a 20°C. Calcule la masa de á

s344n2d4d5 [400]

Answer:

52.45g

Explanation:

The computation of the mass of pure acetic acid in 125mL of this solution is shown below:

The percentage of mass would be equivalent to the g of solute in each 100g of water

As we know that

density = mass ÷ volume

So,

Volume = mass ÷ density

V = 100g / 1.049 (g / ml)

V = 95.328 mL

Now In every 95,328 ml of C_2H_4O_2 there are 40g of C_2H_4O_2

i.e.

each 125ml of C_2H_4O_2 there are 52.45g

SO,

x = 40g. 125ml ÷ 95.328

x = 52.45g

4 0

3 years ago