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3 years ago

Match each vocabulary word with the correct definition.

Chemistry

2 answers:

Nikolay [14]3 years ago

4 0

1. antibody
2. antigen
3. leukocyte
4. phagocytosis
5. dehydration

Anni [7]3 years ago

3 0

Answer:

1. antibody

2. antigen

3. leukocyte

4. phagocytosis

5. dehydration

Explanation:

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24. When the equation

serg [7]

Answer:

Explanation:

Explanation:

The reaction expression is given as:

Al(OH)₃ + HNO₃ → H₂O + Al(NO₃)₃

To solve this problem, let us assign coefficient a,b,c and d to each specie;

aAl(OH)₃ + bHNO₃ → cH₂O + dAl(NO₃)₃

Conserving Al : a = d

O: 3a + 3b = c + 9d

H: 3a + b = 2c

N: b = 3d

let a = 1 , d = 1, b = 3 , c = 3

Multiply through by 3;

a = 1, b = 3, c = 3 and d = 1

Al(OH)₃ + 3HNO₃ → 3H₂O + Al(NO₃)₃

7 0

3 years ago

Read 2 more answers

Mercury has a density of 13.6 grams/mL, what is this density in cg/L?

Firdavs [7]

It would be 4.6cgL not sure tho because I didint do that good in this

4 0

3 years ago

Determine the mass of water in kg, if 4300 cal of energy is placed in water, resulting in a temperature change to 101.0 oC from

Airida [17]

Answer:

0.5059kg

Explanation:

The heat absorbed for the water is determined using the equation:7

Q = C×m×ΔT

Where Q is heat absorbed (4300cal)

C is specific heat (1cal/g°C)

m is the mass in grams

ΔT is change in °C (101.0°C - 92.5°C = 8.5°C)

Replacing:

4300cal = 1cal/g°C×m×8.5°C

505.9g = m

In kg, the mass of water is:

6 0

3 years ago

During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

8 0

4 years ago

A breathing mixture used by deep-sea divers contains helium, oxygen, and carbon dioxide. What is the partial pressure of oxygen

Kay [80]

Answer is: pressure of oxygen is 31,3 kPa.
The total pressure of an ideal gas mixture is the sum of the partial pressures of the gases in the mixture.
p(mixture) = p(helium) + p(oxygen) + p(carbon dioxide).
p(oxygen) = p(mixture) - (p(helium) + p(carbon dioxide)).
p(oxygen) = 101,4 kPa - (68,7 kPa + 1,4 kPa).
p(oxygen) = 101,4 kPa - 70,1 kPa.
p(oxygen) = 31,3 kPa.

3 0

3 years ago