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2 years ago

What does vascular mean

Chemistry

1 answer:

-BARSIC- [3]2 years ago

5 0

<span>of, relating to, affecting, or consisting of a vessel or vessels, especially those that carry blood</span>

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A teacher places a cup of coffee onto an electronic balance at the front of the science laboratory. The teacher then adds three

motikmotik

Answer:

it will be same as before

4 0

2 years ago

Read 2 more answers

Please help

KonstantinChe [14]

Answer: There are 0.5 grams of barium sulfate are present in 250 of 2.0 M $BaSO_{4}$ solution.

Explanation:

Given: Molarity of solution = 2.0 M

Volume of solution = 250 mL

Convert mL int L as follows.

$1 mL = 0.001 L\\250 mL = 250 mL \times \frac{0.001 L}{1 mL}\\= 0.25 L$

Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given $BaSO_{4}$ solution is as follows.

$Molarity = \frac{mass}{Volume (in L)}\\2.0 M = \frac{mass}{0.25 L}\\mass = 0.5 g$

Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M $BaSO_{4}$ solution.

7 0

3 years ago

Given: 4Na + O2 → 2Na2O In this chemical reaction, how many moles of Na2O will be produced if 2.90 moles of Na react completely?

crimeas [40]

Since we already have the balanced equation, we know that the ratio between $Na:Na_{2}O$ is $4:2$ respectively.

So then we can set up a proportion to find the number of moles produced when 2.90 moles of Na react completely:

$\frac{4mol_{Na}}{2mol_{Na_{2}O}} =\frac{2.90mol_{Na}}{xmol_{Na_{2}O}}$

Then we cross multiply and solve for x:

$4x=5.8$

$x=1.45$

Therefore, we know that when 2.90 moles of Na react completely, there are 1.45 moles of $Na_{2}O$ that are produced.

3 0

2 years ago

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given

tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

$K_a$ = $1.34*10^{-5$

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝) $\alpha = \sqrt{\frac{K_a}{C} }$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }$

$\alpha = \sqrt{4.9629*10^{-5}}$

$\alpha = 7.044*10^{-3$

percentage% (∝) = $7.044*10^{-3}*100$

= 0.704%

For Part B; where Concentration of B = $7.84*10^{-2$ M

$\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }$

$\alpha = \sqrt{1.709*10^{-4} }$

$\alpha = 0.0130\\$

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= $1.92*10^{-2} M$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }$

$\alpha = \sqrt{6.802*10^{-4}}$

$\alpha =0.02608$

percentage% (∝) = 0.02608 × 100%

= 2.60%

7 0

3 years ago

I need help with this

Lelechka [254]

Answer:

The answer is its equal to the volume of its container.

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I hope this helps! :)

7 0

3 years ago